Question: Q. 6. A point charge $+Q$ is placed at point $O$ as shown in the figure. Is the potential difference $V_{A}-V_{B}$ positive, negative or zero?
U] [Delhi Set I, II, III 2016]
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Solution:
Ans. Positive.
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Detailed Answer :
Let $O A=r_{A}, O B=r_{B}$
$$ V_{A}-V_{B}=\frac{1}{4 \pi \varepsilon_{0}} \cdot Q\left(\frac{1}{r_{A}}-\frac{1}{r_{B}}\right) $$
Since $\frac{1}{r_{\mathrm{A}}}-\frac{1}{r_{\mathrm{B}}}$ is positive quantity. Hence, $V_{A}-V_{B}$ is positive.
Answering Tip
- If $r_{A}<r_{B}$
Then $\frac{1}{r_{\mathrm{A}}}>\frac{1}{r_{\mathrm{B}}}$
So, $\frac{1}{r_{\mathrm{A}}}-\frac{1}{r_{\mathrm{B}}}=+$ ve
[बI Q. 7. A charge ’ $q$ ’ is moved from a point $A$ above a dipole of dipole moment ’ $p$ ’ to a point $B$ below the dipole in equatorial plane without acceleration. Find the work done in the process.
U [O.D. I, II, III 2016]
Ans. No work is done.
$$ \begin{equation*} W=q V_{A B}=q \times 0=0 \tag{1} \end{equation*} $$
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Detailed Answer :
It is seen that potential due to dipole at any point on equatorial line in equatorial plane is zero. From the figure, points $A & B$ lies in equatorial plane of dipole, where work done in moving charge ’ $q$ ’ from point $A$ to $B$ without acceleration will be zero. 1
