TS EAMCET · Physics · Electrostatics
A cube of side \(L\) has point charges \(+q\) located at its seven vertices and \(-q\) at remaining one vertex. The electric field at its centre is found to be \(|\mathbf{E}|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)\). The magnitude of constant \(\alpha\) is 
- A \(\frac{4}{3}\)
- B \(\frac{8}{3}\)
- C 3
- D 1
Answer & Solution
Correct Answer
(B) \(\frac{8}{3}\)
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure Clearly, \(5 x\)-components of \(\mathbf{E}\) will be left to right, while \(3 x\)-components of \(\mathbf{E}\) will be from right to left. From the figure given below \(r^2=\frac{L^2}{4}+\frac{L^2}{2}=\frac{3 L^2}{4}\) So,…
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