359363
Figure shows three circular arcs, each of radius \({R}\) and total charge as indicated. The net electric potential at the centre of curvature is
359364
Two thin wire rings each having a radius \(R\) are placed at distance d apart with their axes coinciding. The charges on the two rings are \(+Q\) and \(-Q\). The potential difference between the rings is
The potentials at the centre of the rings are \({V_1} = \frac{Q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) and \({V_2} = \frac{{ - Q}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) \(\Delta V = {V_1} - {V_2}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359365
The electric potential at the centre of two concentric half rings of radii \(R_{1}\) and \(R_{2}\), having same linear charge density \(\lambda\) is
1 \(\dfrac{2 \lambda}{\varepsilon_{0}}\)
2 \(\dfrac{\lambda}{2 \varepsilon_{0}}\)
3 \(\dfrac{\lambda}{\varepsilon_{0}}\)
4 \(\dfrac{\lambda}{4 \varepsilon_{0}}\)
Explanation:
Given: \(R_{1}\) and \(R_{2}\) be the radius of ring 1 and 2 respectively. As, linear charge density ' \(\lambda\) ' is same for both the rings, then \(Q_{1}=\lambda\left(\pi R_{1}\right)\) and \(Q_{2}=\lambda\left(\pi R_{2}\right)\) Here, \(Q_{1}\) and \(Q_{2}\) be the charge on ring 1 and ring 2 respectively. From all points on ring 1, distance of point \(O\) is \(R_{1}\) and from all points on ring 2 distance of point \(O\) is \(R_{2}\). Therefore, electric potential at centre ' \(O\) ' is, \(V=V_{1}+V_{2}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{1}}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{2}}{R_{2}}\) \(\Rightarrow V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{1}\right)}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{2}\right)}{R_{2}}\) or \(V=\dfrac{1}{4 \pi \varepsilon_{0}} \cdot \lambda \pi(1+1)\) or \(V=\dfrac{\lambda}{2 \varepsilon_{0}}\) So, correct option is (2)
359363
Figure shows three circular arcs, each of radius \({R}\) and total charge as indicated. The net electric potential at the centre of curvature is
359364
Two thin wire rings each having a radius \(R\) are placed at distance d apart with their axes coinciding. The charges on the two rings are \(+Q\) and \(-Q\). The potential difference between the rings is
The potentials at the centre of the rings are \({V_1} = \frac{Q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) and \({V_2} = \frac{{ - Q}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) \(\Delta V = {V_1} - {V_2}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359365
The electric potential at the centre of two concentric half rings of radii \(R_{1}\) and \(R_{2}\), having same linear charge density \(\lambda\) is
1 \(\dfrac{2 \lambda}{\varepsilon_{0}}\)
2 \(\dfrac{\lambda}{2 \varepsilon_{0}}\)
3 \(\dfrac{\lambda}{\varepsilon_{0}}\)
4 \(\dfrac{\lambda}{4 \varepsilon_{0}}\)
Explanation:
Given: \(R_{1}\) and \(R_{2}\) be the radius of ring 1 and 2 respectively. As, linear charge density ' \(\lambda\) ' is same for both the rings, then \(Q_{1}=\lambda\left(\pi R_{1}\right)\) and \(Q_{2}=\lambda\left(\pi R_{2}\right)\) Here, \(Q_{1}\) and \(Q_{2}\) be the charge on ring 1 and ring 2 respectively. From all points on ring 1, distance of point \(O\) is \(R_{1}\) and from all points on ring 2 distance of point \(O\) is \(R_{2}\). Therefore, electric potential at centre ' \(O\) ' is, \(V=V_{1}+V_{2}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{1}}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{2}}{R_{2}}\) \(\Rightarrow V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{1}\right)}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{2}\right)}{R_{2}}\) or \(V=\dfrac{1}{4 \pi \varepsilon_{0}} \cdot \lambda \pi(1+1)\) or \(V=\dfrac{\lambda}{2 \varepsilon_{0}}\) So, correct option is (2)
359363
Figure shows three circular arcs, each of radius \({R}\) and total charge as indicated. The net electric potential at the centre of curvature is
359364
Two thin wire rings each having a radius \(R\) are placed at distance d apart with their axes coinciding. The charges on the two rings are \(+Q\) and \(-Q\). The potential difference between the rings is
The potentials at the centre of the rings are \({V_1} = \frac{Q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) and \({V_2} = \frac{{ - Q}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) \(\Delta V = {V_1} - {V_2}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359365
The electric potential at the centre of two concentric half rings of radii \(R_{1}\) and \(R_{2}\), having same linear charge density \(\lambda\) is
1 \(\dfrac{2 \lambda}{\varepsilon_{0}}\)
2 \(\dfrac{\lambda}{2 \varepsilon_{0}}\)
3 \(\dfrac{\lambda}{\varepsilon_{0}}\)
4 \(\dfrac{\lambda}{4 \varepsilon_{0}}\)
Explanation:
Given: \(R_{1}\) and \(R_{2}\) be the radius of ring 1 and 2 respectively. As, linear charge density ' \(\lambda\) ' is same for both the rings, then \(Q_{1}=\lambda\left(\pi R_{1}\right)\) and \(Q_{2}=\lambda\left(\pi R_{2}\right)\) Here, \(Q_{1}\) and \(Q_{2}\) be the charge on ring 1 and ring 2 respectively. From all points on ring 1, distance of point \(O\) is \(R_{1}\) and from all points on ring 2 distance of point \(O\) is \(R_{2}\). Therefore, electric potential at centre ' \(O\) ' is, \(V=V_{1}+V_{2}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{1}}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{2}}{R_{2}}\) \(\Rightarrow V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{1}\right)}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{2}\right)}{R_{2}}\) or \(V=\dfrac{1}{4 \pi \varepsilon_{0}} \cdot \lambda \pi(1+1)\) or \(V=\dfrac{\lambda}{2 \varepsilon_{0}}\) So, correct option is (2)
359363
Figure shows three circular arcs, each of radius \({R}\) and total charge as indicated. The net electric potential at the centre of curvature is
359364
Two thin wire rings each having a radius \(R\) are placed at distance d apart with their axes coinciding. The charges on the two rings are \(+Q\) and \(-Q\). The potential difference between the rings is
The potentials at the centre of the rings are \({V_1} = \frac{Q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) and \({V_2} = \frac{{ - Q}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]\) \(\Delta V = {V_1} - {V_2}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359365
The electric potential at the centre of two concentric half rings of radii \(R_{1}\) and \(R_{2}\), having same linear charge density \(\lambda\) is
1 \(\dfrac{2 \lambda}{\varepsilon_{0}}\)
2 \(\dfrac{\lambda}{2 \varepsilon_{0}}\)
3 \(\dfrac{\lambda}{\varepsilon_{0}}\)
4 \(\dfrac{\lambda}{4 \varepsilon_{0}}\)
Explanation:
Given: \(R_{1}\) and \(R_{2}\) be the radius of ring 1 and 2 respectively. As, linear charge density ' \(\lambda\) ' is same for both the rings, then \(Q_{1}=\lambda\left(\pi R_{1}\right)\) and \(Q_{2}=\lambda\left(\pi R_{2}\right)\) Here, \(Q_{1}\) and \(Q_{2}\) be the charge on ring 1 and ring 2 respectively. From all points on ring 1, distance of point \(O\) is \(R_{1}\) and from all points on ring 2 distance of point \(O\) is \(R_{2}\). Therefore, electric potential at centre ' \(O\) ' is, \(V=V_{1}+V_{2}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{1}}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q_{2}}{R_{2}}\) \(\Rightarrow V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{1}\right)}{R_{1}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\lambda\left(\pi R_{2}\right)}{R_{2}}\) or \(V=\dfrac{1}{4 \pi \varepsilon_{0}} \cdot \lambda \pi(1+1)\) or \(V=\dfrac{\lambda}{2 \varepsilon_{0}}\) So, correct option is (2)
