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GUJCET · Physics · Electrostatic Potential and Capacitance
Eight charges, each of magnitude \(q\) are placed at the vertices of a cube placed in vacuum. Electric potential at the centre of the cube due to this system of charges is ________.
( \(\varepsilon_0=\) permittivity of vacuum, \(a=\) length of each side of the cube.)
- A zero
- B \(\frac{\sqrt{3} q}{\pi \varepsilon_0 a}\)
- C \(\frac{2 q}{\pi \varepsilon_0 a}\)
- D \(\frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}\)
Answer & Solution
Correct Answer
(D) \(\frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}\)
Step-by-step Solution
Detailed explanation
(D)
diagonal length of cube \(=\sqrt{3} a\) distance of vertex from centre ' \(O\) ' is \(r\), then
\(r=\frac{\sqrt{3} a}{2}\)
potential at ' \(O\) ' due to electric charges placed on each vertex.
\(\begin{array}{l} V =8 \frac{k q}{r} \\ V=8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{\frac{\sqrt{3} a}{2}} \\ V=\frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}\end{array}\)
diagonal length of cube \(=\sqrt{3} a\) distance of vertex from centre ' \(O\) ' is \(r\), then
\(r=\frac{\sqrt{3} a}{2}\)
potential at ' \(O\) ' due to electric charges placed on each vertex.
\(\begin{array}{l} V =8 \frac{k q}{r} \\ V=8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{\frac{\sqrt{3} a}{2}} \\ V=\frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}\end{array}\)
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