A clock dial has point charges -q, -2q, -3q, | vedclass

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(B) The electric field $E$ due to a charge $Q$ at distance $r$ is given by $E = \frac{k|Q|}{r^2}$. Since all charges are at the same distance $r$ from

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$9$

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(B) The electric field $E$ due to a charge $Q$ at distance $r$ is given by $E = \frac{k|Q|}{r^2}$. Since all charges are at the same distance $r$ from the centre,the electric field due to a charge $nq$ is proportional to $n$. Let the electric field due to charge $-q$ at position $1$ be $\vec{E}_1$ directed towards the charge (since it is negative). The net electric field $\vec{E}_{net}$ is the vector sum of all fields: $\vec{E}_{net} = \sum_{n=1}^{12} \vec{E}_n$. Note that the charge at position $n$ is $-nq$. The field $\vec{E}_n$ is directed from the centre towards the position $n$. We can pair opposite charges: $(n)$ and $(n+6)$. The net field from these two is $\vec{E}_{net, n} = \vec{E}_n + \vec{E}_{n+6} = \frac{k}{r^2} [(-n\hat{r}_n) + (-(n+6)\hat{r}_{n+6})]$. Since $\hat{r}_{n+6} = -\hat{r}_n$,we have $\vec{E}_{net, n} = \frac{k}{r^2} [-n\hat{r}_n + (n+6)\hat{r}_n] = \frac{6k}{r^2} \hat{r}_n$. This means the net field from each pair $(n, n+6)$ is directed towards the position $n+6$. There are $6$ such pairs: $(1,7), (2,8), (3,9), (4,10), (5,11), (6,12)$. The net field is $\vec{E}_{net} = \frac{6k}{r^2} [\hat{r}_7 + \hat{r}_8 + \hat{r}_9 + \hat{r}_{10} + \hat{r}_{11} + \hat{r}_{12}]$. These unit vectors point towards $7, 8, 9, 10, 11, 12$ on the clock. The resultant of these vectors points towards the direction between $9$ and $10$,which corresponds to $9:30$.

Column-$I$	Column-$II$
$(A)$ Ring along its axis	$(P)$ Graph with a peak at a distance $r > 0$
$(B)$ Uniformly charged solid sphere	$(Q)$ Graph increasing linearly for $r < R$ and decreasing as $1/r^2$ for $r > R$
$(C)$ Uniformly charged spherical shell	$(R)$ Graph with zero field for $r < R$ and decreasing as $1/r^2$ for $r > R$
$(D)$ Combination of charge $+Q$ and $-Q$ at the perpendicular bisector	$(S)$ Graph with a maximum at the center and decreasing as $r$ increases

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