JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Easy
- MCQ
Two charges \(\mathrm{q}_1\) and \(\mathrm{q}_2\) are separated by a distance of 30 cm. A third charge \(q_3\) initially at ' C ' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of \(q_3\) from \(C\) to \(D\) is given by \(\frac{\mathrm{q}_3 \mathrm{~K}}{4 \pi \epsilon_0}\), the value of K is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(8 \mathrm{q}_2\)
Step-by-step Solution
Detailed explanation
Potential at C
\(\mathrm{V}_{\mathrm{C}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.5}\)
Potential at D
\(\mathrm{V}_{\mathrm{D}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.1}\)
\(\Delta \mathrm{U}=\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{c}}\right)\left(\mathrm{q}_3\right)=\left(\frac{\mathrm{kq}_2}{0.1}-\frac{\mathrm{kq}_2}{0.5}\right)\left(\mathrm{q}_3\right)\)
\(\Delta \mathrm{U}=8 \mathrm{kq}_2 \mathrm{q}_3=\frac{8 \mathrm{q}_2 \mathrm{q}_3}{4 \pi \varepsilon_0}\)
\(\mathrm{V}_{\mathrm{C}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.5}\)
Potential at D
\(\mathrm{V}_{\mathrm{D}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.1}\)
\(\Delta \mathrm{U}=\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{c}}\right)\left(\mathrm{q}_3\right)=\left(\frac{\mathrm{kq}_2}{0.1}-\frac{\mathrm{kq}_2}{0.5}\right)\left(\mathrm{q}_3\right)\)
\(\Delta \mathrm{U}=8 \mathrm{kq}_2 \mathrm{q}_3=\frac{8 \mathrm{q}_2 \mathrm{q}_3}{4 \pi \varepsilon_0}\)
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