The electric field at a place is given by ove | vedclass

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(C) The work done by a constant electric field $\overrightarrow E$ on a charge $q_0$ moving from point $A$ to $B$ is given by the formula $W = \overri

Vedclass · Accepted Answer

$q_0 a E_0$

Vedclass · Answer

(C) The work done by a constant electric field $\overrightarrow E$ on a charge $q_0$ moving from point $A$ to $B$ is given by the formula $W = \overrightarrow F \cdot \overrightarrow d$,where $\overrightarrow F = q_0 \overrightarrow E$ is the electric force and $\overrightarrow d$ is the displacement vector.The displacement vector $\overrightarrow d$ from point $A(0, a)$ to point $B(a, 0)$ is calculated as:$\overrightarrow d = (x_B - x_A) \widehat i + (y_B - y_A) \widehat j$$\overrightarrow d = (a - 0) \widehat i + (0 - a) \widehat j = a \widehat i - a \widehat j$The electric field is $\overrightarrow E = E_0 \widehat i$.Now,calculate the work done:$W = q_0 \overrightarrow E \cdot \overrightarrow d$$W = q_0 (E_0 \widehat i) \cdot (a \widehat i - a \widehat j)$$W = q_0 E_0 a (\widehat i \cdot \widehat i) - q_0 E_0 a (\widehat i \cdot \widehat j)$Since $\widehat i \cdot \widehat i = 1$ and $\widehat i \cdot \widehat j = 0$,we get:$W = q_0 E_0 a (1) - 0 = q_0 a E_0$.

The electric field at a place is given by $\overrightarrow E = E_0 \widehat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path. Find the work done by the electric field in this motion.

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