Question: Q. 4. $A$ test charge $q$ is moved without acceleration from $A$ to $C$ along the path from $A$ to $B$ and then from $B$ to $C$ in electric field $E$ as shown in the figure.
(i) Calculate the potential difference between $A$ and C.
(ii) At which point (of the two) is the electric potential more and why?
A [O.D. I, II, III 2012]
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Solution:
Ans. Since,
$E=-d V / d r$
$E=\left(V_{C}-V_{A}\right) / 4$
$1 / 2$
Therefore, $\quad V_{A}-V_{C}=-4 E$
At point $C$, potential is more
$1 / 2$
etric field is in the direction in which the potential decreases.
[CBSE Marking Scheme 2012]
Detailed Answer :
(i) From the figure, as work done is independent of path, so moving a charge $q$ from point $A$ to $B$ and further from point $B$ to $C$, we will take a direct path from point $A$ to $C$. So potential difference between points $A$ and $C$ is given as :
$$ \begin{align*} V_{C}-V_{A} & =-E . d l \ & =-\int_{A}^{C} E d l \cos \theta \end{align*} $$
As direction of test charge ’ $q$ ’ is opposite to direction of electric field $E$ along $A C$, so angle between $\vec{E}$ and $\overrightarrow{d l}$ will be $180^{\circ}$, hence we can write this as :
$$ \begin{aligned} V_{C}-V_{A} & =-E . d l \ & =-\int_{A}^{C} E d l \cos \theta \end{aligned} $$
(ii) As direction of electric field is high to low potential, so $V_{C}>V$
Also, $\quad V_{C}-V_{A}=4 E$ with $E$ as positive $\quad 1 / 2$ Potentialincreases when charge is moved against the direction of electric field, so potential will be more at point $C$.
