Question

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is-

• A. $\frac{Q}{2\pi\varepsilon_{0}R}$
• B. $\frac{Q}{4\pi\varepsilon_{0}R}$
• C. $\frac{2Q}{\pi\varepsilon_{0}R}$
• D. $\frac{Q}{\pi\varepsilon_{0}R}$

Answer 1

Answer: (A)

$\frac{Q}{2\pi\varepsilon_{0}R}$

Answer 2

V = V₁ + V₂ + V₃

= $\frac { 1 } { 4 \pi \varepsilon _ { 0 } } \cdot \frac { \mathrm { Q } } { \mathrm { R } } + \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \left[ \frac { - 2 \mathrm { Q } } { \mathrm { R } } \right] + \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \left[ \frac { 3 \mathrm { Q } } { \mathrm { R } } \right]$

=$\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$.