Class 12 NCERT PHYSICS • Chapter 2: NCERT: Electrostatic Potential and Capacitance • Question 1

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QUESTION 1

Two charges $5 \times 10^{-8} C$ and $-3 \times 10^{-8} C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

SOLUTION

We need to determine the positions along the line joining two point charges where the net electric potential is zero.

Concept: Electric potential

The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula:

V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

According to the principle of superposition, the total potential at any point is the algebraic sum of the potentials due to each individual charge:

V_{\text{total}} = V_1 + V_2

Given:

First charge: $q_1 = 5 \times 10^{-8}\,\text{C}$
Second charge: $q_2 = -3 \times 10^{-8}\,\text{C}$
Separation distance: $d = 16\,\text{cm}$
Potential at infinity: $V_{\infty} = 0$

Solving:

Let $x$ be the distance (in $\text{cm}$ ) from the positive charge $q_1$ where the net potential is zero. There are two possible locations on the line joining the charges.

Case 1: The point lies between the two charges ( $0 < x < 16$ ).

The distance from $q_1$ is $x$ .
The distance from $q_2$ is $16 - x$ .
Setting the total potential to zero:

\begin{aligned} \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{x} + \frac{q_2}{16 - x} \right) &= 0 \\ \frac{5 \times 10^{-8}}{x} + \frac{-3 \times 10^{-8}}{16 - x} &= 0 \\ \frac{5}{x} &= \frac{3}{16 - x} \\ 5(16 - x) &= 3x \\ 80 - 5x &= 3x \\ 8x &= 80 \\ x &= 10\,\text{cm} \end{aligned}

Case 2: The point lies outside the charges.

Since the magnitude of $q_1$ is greater than $q_2$ , the zero potential point must be closer to the smaller charge $q_2$ to balance the potentials. Thus, the point lies at $x > 16$ .
The distance from $q_1$ is $x$ .
The distance from $q_2$ is $x - 16$ .
Setting the total potential to zero:

\begin{aligned} \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{x} + \frac{q_2}{x - 16} \right) &= 0 \\ \frac{5 \times 10^{-8}}{x} - \frac{3 \times 10^{-8}}{x - 16} &= 0 \\ \frac{5}{x} &= \frac{3}{x - 16} \\ 5(x - 16) &= 3x \\ 5x - 80 &= 3x \\ 2x &= 80 \\ x &= 40\,\text{cm} \end{aligned}

The electric potential is zero at a distance of $10\,\text{cm}$ and $40\,\text{cm}$ from the positive charge on the line joining the two charges.

10\,\text{cm} \text{ and } 40\,\text{cm} \text{ from the } 5 \times 10^{-8}\,\text{C} \text{ charge}

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Q 2 →

NCERT: Electrostatic Potential and Capacitance

Two charges 5×10−8C5 \times 10^{-8} C5×10−8C and −3×10−8C-3 \times 10^{-8} C−3×10−8C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

SOLUTION

We need to determine the positions along the line joining two point charges where the net electric potential is zero.

Concept: Electric potential

Two charges $5 \times 10^{-8} C$ and $-3 \times 10^{-8} C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.