MHT CET · Physics · Electrostatics
The point charges \(+q,-q,-q,+q,+Q\) and \(-q\) are placed at the vertices of a regular hexagon ABCDEF as shown in figure. The electric field at the centre of hexagon ' \(O\) ' due to the five charges at \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and F is thrice the electric field at centre ' O ' due to charge +Q at E alone. The value of Q is
- A \(\frac{+q}{3}\)
- B \(\frac{\mathrm{qq}}{5}\)
- C \(\frac{+q}{6}\)
- D \(+6 q\).
Answer & Solution
Correct Answer
(A) \(\frac{+q}{3}\)
Step-by-step Solution
Detailed explanation
Electric field at ' \(O\) ' due to charge ' \(Q\) ' is given by, \(E=\frac{k Q}{r^2}\)
Electric field at ' \(O\) ' due to charges \(+q\) at \(A\) and \(D\) are equal and opposite and hence they will cancel each other. Similarly electric field at ' O ' due to charges -q at F and C will also cancel each other.
Electric field at ' O ' due to charge -q at B is given by,
\(E^{\prime}=\frac{k q}{r^2}\)
Now, \(\mathrm{E}^{\prime}=3 \mathrm{E}\)
\(\therefore \quad \frac{\mathrm{kq}}{\mathrm{r}^2}=\frac{3 \mathrm{kQ}}{\mathrm{r}^2} \Rightarrow \mathrm{Q}=\frac{\mathrm{q}}{3}\)
Electric field at ' \(O\) ' due to charges \(+q\) at \(A\) and \(D\) are equal and opposite and hence they will cancel each other. Similarly electric field at ' O ' due to charges -q at F and C will also cancel each other.
Electric field at ' O ' due to charge -q at B is given by,
\(E^{\prime}=\frac{k q}{r^2}\)
Now, \(\mathrm{E}^{\prime}=3 \mathrm{E}\)
\(\therefore \quad \frac{\mathrm{kq}}{\mathrm{r}^2}=\frac{3 \mathrm{kQ}}{\mathrm{r}^2} \Rightarrow \mathrm{Q}=\frac{\mathrm{q}}{3}\)
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