Comprehensive Problems on Electric Charges and Coulomb's Law

Effects of Electric Charge on Soap Bubbles

When a soap bubble is given an electric charge, whether that charge is positive or negative, its radius will increase. This phenomenon occurs because the charge distributes itself over the surface of the bubble. Since like charges repel one another, an electrostatic outward force (electrostatic pressure) is generated across the surface. This outward force acts in opposition to the surface tension of the soap film, which typically attempts to minimize the surface area. Consequently, the equilibrium state of the bubble shifts, and the soap bubble expands to a larger radius. Therefore, if a soap bubble is given a negative charge, its radius will increase.

Quantization and Possibility of Electric Charges

The fundamental principle governing the existence of specific charge values is the quantization of charge, which states that the total charge QQ of a body is always an integral multiple of the basic unit of charge ee. This is expressed by the formula Q=neQ = n e, where nn is an integer (n=1,2,3,...n = 1, 2, 3, \text{...}) and ee is the elementary charge, approximately 1.6×10−19 C1.6 \times 10^{-19} \text{ C}.

To determine if a charge is possible, one must divide the given charge by the elementary charge and check for an integer result:

  1. For a charge of 1.6×10−18 C1.6 \times 10^{-18} \text{ C}, the value of nn is 1.6×10−18 C1.6×10−19 C=10\frac{1.6 \times 10^{-18} \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 10. Since 10 is an integer, this charge is possible.

  2. For a charge of 1.6×10−19 C1.6 \times 10^{-19} \text{ C}, the value of nn is 11, which is an integer, making it possible.

  3. For a charge of 1.6×10−20 C1.6 \times 10^{-20} \text{ C}, the value of nn is 1.6×10−20 C1.6×10−19 C=0.1\frac{1.6 \times 10^{-20} \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 0.1. Since 0.1 is not an integer, this charge is not possible.

  4. For a charge of 1.6 C1.6 \text{ C}, the value of nn is 1.6 C1.6×10−19 C=1019\frac{1.6 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 10^{19}. This is a large integer, so the charge is possible.

Thus, a charge of 1.6×10−20 C1.6 \times 10^{-20} \text{ C} is physically impossible.

Electrostatic Interactions Between Multiple Bodies

Consider five balls numbered 1 to 5 suspended by separate threads. The interactions are as follows: pairs (1,2), (2,4), and (4,1) show electrostatic attraction, while pairs (2,3) and (4,5) show repulsion.

From the repulsion observations, we can conclude that balls 2 and 3 carry the same type of charge, and balls 4 and 5 carry the same type of charge. Let us assume ball 2 is positively charged; this means ball 3 is also positively charged. Because pair (2,4) attracts, ball 4 must be negatively charged (since 2 is positive). Because pair (4,5) repels, ball 5 must also be negatively charged.

Now we analyze ball 1. It attracts ball 2 (which is positive) and it also attracts ball 4 (which is negative). If ball 1 were positively charged, it would repel ball 2. If it were negatively charged, it would repel ball 4. The only way ball 1 can attract both positive and negative charges is if ball 1 is neutral. This occurs because a charged object can attract a neutral conductor through the process of electrostatic induction. Therefore, ball 1 must be neutral.

Charge Distribution on Solid and Hollow Conductors

In the case of metallic conductors, such as copper spheres, any excess charge resides entirely on the outer surface of the conductor to minimize potential energy through mutual repulsion. If a solid copper sphere and a hollow copper sphere have the same radius, they have the same external surface area.

The limit to how much charge a sphere can hold is typically determined by the dielectric strength of the surrounding medium (usually air). The electric field at the surface of a sphere of radius RR and charge QQ is given by E = \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{Q}{R^2}. Since both spheres have the same radius RR, they will reach the breakdown electric field of air at the exact same value of QQ. Consequently, both spheres will hold an equal amount of charge.

Mass Changes During the Charging Process

Charging a metallic object involves the physical transfer of electrons, which possess a finite rest mass of approximately 9.11×10−31 kg9.11 \times 10^{-31} \text{ kg}. When sphere A is given a positive charge, it loses electrons, resulting in a decrease in its total mass. Conversely, when sphere B is given an equal amount of negative charge, it must gain a specific number of electrons. The addition of these electrons causes the mass of sphere B to increase. Thus, if spheres A and B are identical and carry equal but opposite charges, the mass of sphere B will be greater than the mass of sphere A.

Equilibrium of Electrostatic and Gravitational Forces

Two particles, each with mass mm and charge qq, are placed at a distance of 16 cm16 \text{ cm}. If they do not experience any net force, it implies that the electrostatic repulsive force (FeF_e) is exactly balanced by the gravitational attractive force (FgF_g).

The magnitudes of these forces are defined as: F_e = \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{q^2}{r^2} Fg=Gm2r2F_g = G \frac{m^2}{r^2}

Setting Fe=FgF_e = F_g yields: \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{q^2}{r^2} = G \frac{m^2}{r^2} \frac{q^2}{m^2} = 4\text{\pi}\text{\epsilon}_0 G \frac{q}{m} = \text{\sqrt{4\text{\pi}\text{\epsilon}_0 G}}

This ratio ensures that the particles remain in a state of neutral equilibrium regardless of the distance rr between them.

Resultant Force on a Charge in a Triangular Configuration

Consider an equilateral triangle ABC with side length L=10 cm=0.1 mL = 10 \text{ cm} = 0.1 \text{ m}. Charges are placed at the vertices as follows: qA=+1 μ Cq_A = +1 \text{ μ C}, qB=−1 μ Cq_B = -1 \text{ μ C}, and qC=+2 μ Cq_C = +2 \text{ μ C}. To find the resultant force on the charge at C, we calculate the individual forces from A and B.

The magnitude of the force from A on C (FACF_{AC}) is: F_{AC} = \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{q_A q_C}{L^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{(0.1)^2} = \frac{18 \times 10^{-3}}{10^{-2}} = 1.8 \text{ N} Since both charges are positive, this force is repulsive, acting along the line extending from A to C.

The magnitude of the force from B on C (FBCF_{BC}) is: F_{BC} = \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{|q_B| q_C}{L^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{(0.1)^2} = 1.8 \text{ N} Since $q_B$ is negative and $q_C$ is positive, this force is attractive, acting along the line from C toward B.

In an equilateral triangle, the internal angle is 60∘60^\text{∘}. The angle between the repulsive force vector FAC\mathbf{F}_{AC} and the attractive force vector FBC\mathbf{F}_{BC} at point C is 120∘120^\text{∘}. The resultant force FRF_R is calculated using the law of cosines for vectors: F_R = \text{\sqrt{F_{AC}^2 + F_{BC}^2 + 2 F_{AC} F_{BC} \text{\cos}(120^\text{∘})}} F_R = \text{\sqrt{(1.8)^2 + (1.8)^2 + 2(1.8)(1.8)(-0.5)}} F_R = \text{\sqrt{1.8^2 + 1.8^2 - 1.8^2}} = 1.8 \text{ N}

Vector Components of Electrostatic Force

Three charges −q1-q_1, +q2+q_2, and −q3-q_3 are positioned in a coordinate system. Charge −q1-q_1 is at the origin (0,0)(0,0). Charge +q2+q_2 is located on the X-axis at a distance bb. Charge −q3-q_3 is located at a distance aa from the origin, such that the line connecting −q3-q_3 to −q1-q_1 makes an angle \text{\theta} with the Y-axis.

We calculate the X-component of the net force acting on −q1-q_1:

  1. Force from q2q_2 on q1q_1 (F12F_{12}): This is an attractive force because the charges are of opposite signs (−q1-q_1 and +q2+q_2). It pulls q1q_1 in the +X+X direction. The magnitude is proportional to q2b2\frac{q_2}{b^2}.

  2. Force from q3q_3 on q1q_1 (F13F_{13}): This is a repulsive force because both charges are negative. The force on q1q_1 acts in the direction opposite to the position of q3q_3. If q3q_3 is at an angle \text{\theta} from the Y-axis (in the positive X-direction), its X-coordinate relative to the origin is a \text{\sin}(\text{\theta}). The repulsive force pushes q1q_1 in the −X-X direction. The X-component of this repulsive force is proportional to \frac{q_3}{a^2} \text{\sin}(\text{\theta}).

The total X-component of the force on −q1-q_1 is the sum of these components: F_x \text{\propto} (\frac{q_2}{b^2} - \frac{q_3}{a^2} \text{\sin}(\text{\theta})).

Electrostatic Equilibrium in Gravity-Free Space

Two small balls, each carrying an equal positive charge QQ, are suspended by two insulated strings of equal length LL from a fixed hook. When this setup is placed in a satellite in a state of weightlessness (zero gravity), the only significant force acting on the balls is the mutual electrostatic repulsion between them.

Because there is no downward gravitational force to pull the balls toward each other, the electrostatic repulsion will push the balls as far apart as the strings allow. This results in the strings becoming collinear, forming an angle of 180∘180^\text{∘} between them. In this configuration, the distance between the two charges is 2L2L.

The tension (TT) in the strings is equal to the repulsive electrostatic force (FeF_e) between the two charges: T = F_e = \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{Q^2}{(2L)^2}

Thus, the angle is 180∘180^\text{∘} and the tension is \frac{1}{4\text{\pi}\text{\epsilon}_0} \frac{Q^2}{4L^2}.

  1. Charge on a Soap Bubble
    If a soap bubble has a radius of $2 \, cm$ and is given a positive charge of $4.8 \times 10^{-6} \, C$, calculate the increase in radius due to the electrostatic pressure.

    • Use the formula for electrostatic pressure:
      P=14πϵ0Q2R4P = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{R^4}
      Where:

    • $\epsilon_0 \approx 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)$

  2. Quantization of Charge
    Determine if the following charges are physically possible:

    • a) $3.20 \times 10^{-19} \, C$

    • b) $2.00 \times 10^{-20} \, C$

    • c) $1.44 \, C$

    For each charge, divide by the elementary charge $e = 1.6 \times 10^{-19} \, C$ and check if the result is an integer.

  3. Force between Charged Balls
    Calculate the electrostatic force between two balls, each with a charge of $+5 \mu C$ placed $10 \, cm$ apart.

    • Use Coulomb's law:
      F=kq<em>1q</em>2r2F = k \frac{q<em>1 q</em>2}{r^2}
      Where:

    • $k \approx 9 \times 10^9 \, Nm^2/C^2$

  4. Resultant Force in a Triangular Configuration
    Consider three charges placed at the vertices of an equilateral triangle of side length $L = 0.2 \, m$.

    • $qA = +2 \, \mu C$, $qB = -2 \, \mu C$, and $q_C = +3 \, \mu C$.

    • Calculate the resultant force on charge $q_C$.

  5. Electrostatic Equilibrium in Zero Gravity
    Two balls each with charge $Q = +1 \, \mu C$ are $2 \, m$ apart in a weightless environment.

    • Calculate the electrostatic force between them and describe how it affects their motion.