Q.(a) Calculate the potential at a point P due to a charge of $4 \times 10^{-7}\ \text{C}$ located $9\ \text{cm}$ away.

Concept: Electric potential due to a point charge — the work done per unit charge to bring a test charge from infinity to that point.

(a) The potential at distance $r$ from a point charge $Q$ is

Given $Q = 4 \times 10^{-7}\ \text{C}$, $r = 9\ \text{cm} = 0.09\ \text{m}$, and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$,

(b) Work done $W = qV$, where $q = 2 \times 10^{-9}\ \text{C}$:

$W = (2 \times 10^{-9})(4 \times 10^4) = 8 \times 10^{-5}\ \text{J}$ …

The Concept: Electric Potential

Electric potential is a scalar quantity that tells us the potential energy per unit charge at a point in an electric field. Think of it like height in a gravitational field — a ball at a higher point has more gravitational potential energy. Similarly, a charge at a higher electric potential has more electric potential energy.

For a point charge $Q$, the potential at a distance $r$ is given by:

This formula comes from integrating the electric field from infinity to that point. The constant $k = 1/(4\pi\varepsilon_0) = 9 \times 10^9\ \text{N m}^2/\text{C}^2$.

Step-by-Step Solution

Part (a): Potential at point P

1. Identify the given data

• Charge $Q = 4 \times 10^{-7}\ \text{C}$
• Distance $r = 9\ \text{cm} = 9 \times 10^{-2}\ \text{m}$
• Constant $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$

2. Apply the formula for potential

3. Simplify step by step

First, cancel the 9s:

Combine the powers of 10:

So the potential at P is $4 \times 10^4$ volts.

Part (b): Work done to bring a charge from infinity

4. Understand the relationship

Work done by an external agent to bring a charge $q$ from infinity to a point at potential $V$ is:

This is because potential is defined as work per unit charge: $V = W/q$.

5. Plug in the values …

Method: Potential of a Point Charge, Then Work to Bring a Second Charge to It

This method applies to any two-part problem that first asks for the potential at a point due to a single point charge, and then the work done to bring another charge to that point.

Steps

Step 1: Identify the source charge and the distance to the point

Write down the source charge $q$ (with its sign) and its distance $r$ from the point of interest. Convert every length to metres and every charge to coulombs before substituting — this is the single most common source of error.

Step 2: Find the potential using the point-charge formula

Remember $V$ is a scalar: only the sign and magnitude of $q$ matter, never a direction.

Step 3: Use $W = q'V$ to find the work done bringing a second charge …