Question
The four capacitors, each of $$25\mu F$$ are connected as shown in figure. The $$DC$$ voltmeter reads $$200\,V.$$ The charge on each plate of capacitor is
The four capacitors, each of $$25\mu F$$ are connected as shown in figure. The $$DC$$ voltmeter reads $$200\,V.$$ The charge on each plate of capacitor is
A.
$$ \pm 2 \times {10^{ - 3}}C$$
B.
$$ \pm 5 \times {10^{ - 3}}C$$
C.
$$ \pm 2 \times {10^{ - 2}}C$$
D.
$$ \pm 5 \times {10^{ - 2}}C$$
Answer :
$$ \pm 5 \times {10^{ - 3}}C$$
Solution :
As from the given diagram, potential difference across each capacitor is $$200\,V.$$
So, charge on each plate of capacitor is given by
$$\eqalign{ & Q = \pm CV\,\,\left[ {_{V = \,\,{\text{voltage or potential difference across the capactor}}}^{C = \,\,{\text{capacitance of capacitor}}}} \right] \cr & = \pm 25 \times {10^{ - 6}} \times 200 = \pm 5 \times {10^{ - 3}}C \cr} $$
As from the given diagram, potential difference across each capacitor is $$200\,V.$$
So, charge on each plate of capacitor is given by
$$\eqalign{ & Q = \pm CV\,\,\left[ {_{V = \,\,{\text{voltage or potential difference across the capactor}}}^{C = \,\,{\text{capacitance of capacitor}}}} \right] \cr & = \pm 25 \times {10^{ - 6}} \times 200 = \pm 5 \times {10^{ - 3}}C \cr} $$
