Problem 144 - A Complete Resource Book in Physics for JEE Main

Question

Electric potential \(V\) in volt in a region is given by \(V=a x^{2}+a y^{2}+2 a z^{2}\), where \(a\) is a constant. Work done by the field when a \(2 \mu \mathrm{C}\) charge moves from point \((0,0,0.1 \mathrm{~m})\) to origin is \(5 \times 10^{-5} \mathrm{~J}\). Find \(a ?\) (in \(\mathrm{V} / \mathrm{m}^{2}\) )

Step-by-Step Solution

Answer

The constant 'a' in the electric potential equation is \(1.25 \times 10^{3} \, \mathrm{V} / \mathrm{m}^{2}\).

Step 1: Write down the given information

We have the electric potential formula as: \(V = ax^2 + ay^2 + 2az^2\), and we know: Charge moved, \(q = 2 \times 10^{-6} \, C\), Initial point, \((x_1, y_1, z_1) = (0, 0, 0.1) \, m\), Final point, \((x_2, y_2, z_2) = (0, 0, 0)\), Work done, \(W = 5 \times 10^{-5} \, J\). Our goal is to find the value of 'a' in volts per square meter (\(\mathrm{V}/\mathrm{m}^2\)).

Step 2: Calculate the electric potential difference

When the charge moves from the initial point to the final point, there will be a change in electric potential. Let's denote the initial potential as \(V_1\) and the final potential as \(V_2\). We can write the electric potentials at both points using the given formula: \(V_1 = a(0)^2 + a(0)^2 + 2a(0.1)^2\), \(V_2 = a(0)^2 + a(0)^2 + 2a(0)^2\). Now, calculating the potential difference \(\Delta V = V_2 - V_1\): \(\Delta V = a(0 - 2(0.1)^2) = -0.02a\).

Step 3: Use the work-energy theorem to derive work done by electric field

The relation between the work done by the electric force and the potential difference is given by \(W = -q\Delta V\), where 'W' is the work done by the electric field, 'q' is the charge and \(\Delta V\) is the electric potential difference. Using the given values for W and q, and the expression for \(\Delta V\) from Step 2, we get: \(5 \times 10^{-5} = - (2 \times 10^{-6})(-0.02a)\).

Step 4: Solve for 'a'

Now, we will solve the equation for 'a': \(5 \times 10^{-5} = 4 \times 10^{-8}a\). Dividing both sides by \(4 \times 10^{-8}\) gives: \(a = \frac{5 \times 10^{-5}}{4 \times 10^{-8}}\). Now, doing the division, we find: \(a = 1.25 \times 10^{3} \, \mathrm{V} / \mathrm{m}^{2}\). So, the constant 'a' in the electric potential equation is \(1.25 \times 10^{3} \, \mathrm{V} / \mathrm{m}^{2}\).