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Question

(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 = \overr

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 $\overrightarrow{d} = \overrightarrow{r}_B - \overrightarrow{r}_A$ is the displacement vector.Given $\overrightarrow{E} = E_0 \hat{i}$ and the coordinates of the points are $A(0, a)$ and $B(a, 0)$.The displacement vector $\overrightarrow{d} = (a - 0) \hat{i} + (0 - a) \hat{j} = a \hat{i} - a \hat{j}$.Now,calculate the work done:$W = (q_0 \overrightarrow{E}) \cdot \overrightarrow{d}$$W = (q_0 E_0 \hat{i}) \cdot (a \hat{i} - a \hat{j})$$W = q_0 E_0 a (\hat{i} \cdot \hat{i}) - q_0 E_0 a (\hat{i} \cdot \hat{j})$Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$,we get:$W = q_0 E_0 a (1) - 0 = q_0 a E_0$.

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

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