Question:
Two large circular metal plates each of radius \(50\,\text{cm}\) carrying equal and unlike charges are parallel to each other. If the electric field between the plates is
\[
720\ \text{N C}^{-1},
\]
then the magnitude of charge on any one plate is
Two large circular metal plates each of radius \(50\,\text{cm}\) carrying equal and unlike charges are parallel to each other. If the electric field between the plates is
\[
720\ \text{N C}^{-1},
\]
then the magnitude of charge on any one plate is
Show Hint
For two oppositely charged large parallel plates,
\[
E=\frac{\sigma}{\varepsilon_0}.
\]
First find
\[
\sigma=\varepsilon_0E,
\]
then use
\[
Q=\sigma A.
\]
Updated On: Jul 29, 2026
- \(7.5\,\text{nC}\)
- \(10\,\text{nC}\)
- \(2.5\,\text{nC}\)
- \(5\,\text{nC}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For two large oppositely charged parallel plates, the electric field between them is
\[
E=\frac{\sigma}{\varepsilon_0},
\]
where
\[
\sigma=\frac{Q}{A}
\]
is the surface charge density.
Step 1: Calculate the surface charge density. Given, \[ E=720\ \text{N C}^{-1}, \qquad \varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}. \] Therefore, \[ \sigma=\varepsilon_0E. \] \[ =(8.85\times10^{-12})(720). \] \[ =6.372\times10^{-9}\ \text{C m}^{-2}. \]
Step 2: Calculate the area of one plate. Radius, \[ r=50\,\text{cm}=0.5\,\text{m}. \] Hence, \[ A=\pi r^2. \] \[ =\pi(0.5)^2. \] \[ =\frac{\pi}{4} \approx0.785\ \text{m}^2. \]
Step 3: Find the charge on one plate. \[ Q=\sigma A. \] \[ =(6.372\times10^{-9})(0.785). \] \[ \approx5.0\times10^{-9}\ \text{C}. \] \[ Q=5\,\text{nC}. \] Therefore, \[ \boxed{Q=5\times10^{-9}\ \text{C}=5\,\text{nC}} \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: Calculate the surface charge density. Given, \[ E=720\ \text{N C}^{-1}, \qquad \varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}. \] Therefore, \[ \sigma=\varepsilon_0E. \] \[ =(8.85\times10^{-12})(720). \] \[ =6.372\times10^{-9}\ \text{C m}^{-2}. \]
Step 2: Calculate the area of one plate. Radius, \[ r=50\,\text{cm}=0.5\,\text{m}. \] Hence, \[ A=\pi r^2. \] \[ =\pi(0.5)^2. \] \[ =\frac{\pi}{4} \approx0.785\ \text{m}^2. \]
Step 3: Find the charge on one plate. \[ Q=\sigma A. \] \[ =(6.372\times10^{-9})(0.785). \] \[ \approx5.0\times10^{-9}\ \text{C}. \] \[ Q=5\,\text{nC}. \] Therefore, \[ \boxed{Q=5\times10^{-9}\ \text{C}=5\,\text{nC}} \] \[ \boxed{\text{Answer = (D)}} \]
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