Question

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction

Short answer

The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).

Step-by-step solution

Compute the derivative of V(x) with respect to x

To calculate the derivative of V(x), we can use the power rule. We get: \[\frac{dV}{dx} = \frac{d}{dx}[\frac{20}{x^2-4}]\] First, we rewrite V(x) as a power of x: \(20(x^2-4)^{-1}\). Then, applying the power rule: \[\frac{dV}{dx}=-20(-1)(x^2-4)^{-2}(2x)\] Simplifying, we get: \[\frac{dV}{dx}=\frac{40x}{(x^2-4)^2}\]

Find the electric field E(x) at \(x = 4 \mu m\)

Now we have the derivative of V(x), which gives us the electric field E(x): \[E(x)=\frac{40x}{(x^2-4)^2}\] We want to calculate the electric field at \(x = 4 \mu m\). So we substitute the given value: \[E(4) = \frac{40(4)}{(4^2-4)^2}\]

Determine the direction of the electric field

Since E(x) has a positive value for \(x = 4 \mu m\), the electric field's direction points towards the positive x-direction. Now we compute the value of E(4): \[E(4) = \frac{40(4)}{(4^2-4)^2} = \frac{40(4)}{(16-4)^2} = \frac{40(4)}{12^2} = \frac{10}{9} V\mu m^{-1}\] The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).