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GUJCET · Physics · Electric Charges and Fields
Charge of \(1 \mu C\) each is placed on the five corners of a regular hexagon of side 1 m . The electric field at its centre is _________ N/C.
- A \(\frac{6}{5} \times 10^{-6} k\)
- B \(\frac{5}{6} \times 10^{-6} k\)
- C \(5 \times 10^{-6} k\)
- D \(10^{-6} k\)
Answer & Solution
Correct Answer
(D) \(10^{-6} k\)
Step-by-step Solution
Detailed explanation
(D)
As shown in figure, \(E_A\) and \(E_D\) are cancelled (equal in magnitude and opposite in direction).
Similarly \(E _{ B }\) and \(E _{ E }\) are cancelled. The net electric field will be due to the charge \((q=1 \mu C )\) at C only
\(
\begin{array}{l}
E_{C}=\frac{k q}{r^2} \\
E_{C}=\frac{k \times 1 \times 10^{-6}}{(1)^2} \\
E_{C}=10^{-6} k
\end{array}
\)
As shown in figure, \(E_A\) and \(E_D\) are cancelled (equal in magnitude and opposite in direction).
Similarly \(E _{ B }\) and \(E _{ E }\) are cancelled. The net electric field will be due to the charge \((q=1 \mu C )\) at C only
\(
\begin{array}{l}
E_{C}=\frac{k q}{r^2} \\
E_{C}=\frac{k \times 1 \times 10^{-6}}{(1)^2} \\
E_{C}=10^{-6} k
\end{array}
\)
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