Eleven equal point charges, all of them having a charge $+Q$, are placed at all the hour positions of a circular clock of radius $r$. except at the $10 \mathrm{h}$ position. What is the clectric field strength at the centre of the clock?

• Correct answer
  A

  $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark 10
• B

  $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the mark 10 towards the centre
• C

  $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark 6
• D

  Zero.

Solution

The clock diagram is as shown below In the above diagram, charge $+Q$ is not placed at $10 \mathrm{h}$ position. So, net electric ficld strength at centre $C$. $$ \begin{array}{r} \mathrm{E}_{\text {net }}=\mathrm{E}_{1}+\mathrm{E}_{2}+\mathrm{E}_{3}+\mathrm{E}_{4}+\mathrm{E}_{5}+\mathrm{E}_{6} \\ +\mathrm{E}_{7}+\mathrm{E}_{8}+\mathrm{E}_{9}+\mathrm{E}_{11}+\mathrm{E}_{12} \\ \Rightarrow \mathrm{E}_{\text {net }}=\mathrm{E}_{1}+\mathrm{E}_{2}+\mathrm{E}_{3}+\mathrm{E}_{4} \\ +\mathrm{E}_{5}+\mathrm{E}_{6}+\left(-\mathrm{E}_{1}-\mathrm{E}_{2}-\mathrm{E}_{3}-\mathrm{E}_{5}-\mathrm{E}_{6}\right) \end{array} $$ $\Rightarrow \mathrm{E}_{\mathrm{net}}=\mathrm{E}_{4}=\frac{Q}{4 \pi \varepsilon_{0} r^{2}},$ from centre towards the mark $10 .$

Hence, the correct answer is option: 'A'.