Question

In the figure shown, a potential of \[ + 12V\] is given to point $A$ and point $B$ is earthed. What is the potential at point $P$ ?

Answer 1

Hint: In order to find Potential at point P, we need to find the actual potential difference between Point A and point P which is the potential across the capacitor of capacitance $3\mu F$ . Then we will use the general potential difference formula which is ${V_{AP}} = {V_A} - {V_P}$ .

Formula Used:
In case when two capacitors of capacitance ${C_1}$ and ${C_2}$ are connected in parallel combination the net capacitance is calculated as ${C_{net}} = {C_1} + {C_2}$ while, if two capacitors are connected in series ten their net capacitance is calculated as $\dfrac{1}{{{C_{net}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$ .

Complete step by step answer:
Firstly, let us find the equivalent capacitance between points A and B. Since, the capacitors of capacitance $2\mu F$ and $4\mu F$ are connected in parallel with each other so, their resultant capacitance is calculated using formula ${C_{net}} = {C_1} + {C_2}$ so, and we have
${C_{2,4}} = 2 + 4$
$\Rightarrow {C_{2,4}} = 6\mu F$

Now, this net capacitance of ${C_{2,4}} = 6\mu F$ is connected in series with the capacitor of capacitance $3\mu F$ so, the net capacitance between point A and B can be calculated by using formula $\dfrac{1}{{{C_{net}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
so, we have
$\dfrac{1}{{{C_{3,6}}}} = \dfrac{1}{3} + \dfrac{1}{6}$
$\Rightarrow \dfrac{1}{{{C_{3,6}}}} = \dfrac{9}{{18}}$
$\Rightarrow {C_{3,6}} = 2\mu F$
The net capacitance between point A and point B is $2\mu F$ .

Now, let ${Q_{net}}$ be the charge on this net capacitance and the potential at point is given as ${V_A} = + 12V$ and point B is earthed so ${V_B} = 0$ using the formula $Q = CV$ we can write this for net capacitance of $2\mu F$ as
${Q_{net}} = {C_{net}}{V_{AB}}$
$\Rightarrow {Q_{net}} = 2({V_A} - {V_B}) \to (i)$

Now, since capacitor of capacitance $3\mu F$ and capacitor of capacitance ${C_{2,4}} = 6\mu F$ are connected in series then, total charge ${Q_{net}}$ will distribute same across both the capacitors, hence we have the charge stored on $3\mu F$ capacitor is ${Q_{net}}$ and the potential difference across $3\mu F$ is denoted as ${V_{AP}}$ so we can write as:
${Q_{net}} = C{V_{AP}}$
$\Rightarrow {Q_{net}} = 3({V_A} - {V_P}) \to (ii)$
Compare the equations $(i)and(ii)$ we get,
$2({V_A} - {V_B}) = 3({V_A} - {V_P})$
And we have, ${V_A} = + 12V$ and ${V_B} = 0$ hence,
$3{V_P} = 12$
$\therefore {V_P} = 4V$

Hence, the potential at point $P$ is $4\,V$.

Note: It should be remembered that, when two capacitors are connected in series, the same charge flows across them while when two capacitors are connected in parallel, voltage across both the capacitors are the same. When any point of the electrical circuit is grounded the voltage or potential at the point becomes zero.