KCET · Physics · Electrostatics
Two metal plates are separated by \( 2 \mathrm{~cm} \). The potentials of the plates are \( -10 \mathrm{~V} \) and \( +30 \mathrm{~V} \).
The electric field between the two plates is :
- A \( 200 \mathrm{~V} / \mathrm{m} \)
- B \( 500 \mathrm{~V} / \mathrm{m} \)
- C \( 3000 \mathrm{~V} / \mathrm{m} \)
- D \( 1000 \mathrm{~V} / \mathrm{m} \)
Answer & Solution
Correct Answer
(A) \( 200 \mathrm{~V} / \mathrm{m} \)
Step-by-step Solution
Detailed explanation
(A)
\( E=\frac{V}{\alpha}=\frac{30-(-10)}{2 \times 10^{-2}}=2000 \mathrm{Vm}^{-1} \)
( \( \alpha= \) distance between two plates)
\( E=\frac{V}{\alpha}=\frac{30-(-10)}{2 \times 10^{-2}}=2000 \mathrm{Vm}^{-1} \)
( \( \alpha= \) distance between two plates)
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