JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is \(V\). The potential difference between the points A and B (shown in the figure) is :
- A \(\frac{1}{4} \mathrm{~V}\)
- B \(\frac{2}{5} \mathrm{~V}\)
- C \(\frac{3}{4} \mathrm{~V}\)
- D 1 V
Answer & Solution
Correct Answer
(B) \(\frac{2}{5} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Using } \Delta V=E(\Delta d) \\ & V=E(10) \\ & V_{A B}=E \cdot 4=\frac{V}{10} \times 4=\frac{2 V}{5}\end{aligned}\)
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