Problem 65 - A Complete Resource Book in Physics for JEE Main

Question

The electric potential \(V\) (in volt) varies with \(x\) (in metre) according to the relation \(V=5+4 x^{2}\). The force experienced by a negative charge of \(2 \times 10^{-6} \mathrm{C}\) located at \(x=0.5 \mathrm{~m}\) is (A) \(2 \times 10^{-6} \mathrm{~N}\) (B) \(4 \times 10^{-6} \mathrm{~N}\) (C) \(6 \times 10^{-6} \mathrm{~N}\) (D) \(8 \times 10^{-6} \mathrm{~N}\)

Step-by-Step Solution

Answer

The force experienced by the negative charge at x = 0.5m is \(8 \times 10^{-6} N\), which corresponds to option (D).

Step 1: Find the electric field (E(x)) from the potential (V(x))

To find the electric field (E(x)), we'll need to calculate the derivative of the potential with respect to the position (x). The formula for this is: \[ E(x) = - \frac{dV(x)}{dx} \] So first, let's find the derivative of V(x): \[ V(x) = 5 + 4x^2 \] \[ \frac{dV(x)}{dx} = 8x \] Now, calculate the electric field (E(x)): \[ E(x) = - \frac{dV(x)}{dx} = -8x \]

Step 2: Calculate the electric field at x = 0.5m

Now that we have the electric field as a function of position, we can find the electric field at the specific location (x = 0.5m) where the charge is located. \[ E(0.5) = -8(0.5) = -4 \, V/m \]

Step 3: Calculate the force on the negative charge

Finally, we can find the force experienced by the negative charge using the formula F = qE, where F is the force, q is the charge (2 x 10^{-6} C), and E is the electric field at the location of the charge. \[ F = qE = (2 \times 10^{-6} \, C)(-4 \, V/m) \] The negative sign indicates the force is in the opposite direction of the electric field. Since we only need the magnitude of the force, we can ignore the negative sign: \[ F = - (2 \times 10^{-6} \, C)(-4 \, V/m) = 8 \times 10^{-6} \, N \] So the force experienced by the negative charge at x = 0.5m is \(8 \times 10^{-6} N\), which corresponds to option (D).