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Physics 4 Marks Questions

Electrostatic Potential and Capacitance · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 2 questions.

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Question 14 Marks
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply? S
Answer
(a) Given: $C _1= C _2= C _3$ = 9 pF. Total capacitance of the combination in series CS= ?
Image
We know that: $\quad \frac{1}{ C _s}=\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}$
$=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}$
or $C _s=\frac{9}{3}=3 pF$
(b) Let the potentials of capacitors be V1,V2 and V3 respectively = ?
Sum of $V_1, V_2$ and $V_3$
$V_1+V_2+V_3=120 V$
Since$\quad$$\quad$$C _1= C _2= C _3$
$\therefore \quad V _1= V _2= V _3= V \quad$ (assume)
$\therefore \quad V + V + V =120$
$3 V=120$
$V =\frac{120}{3}=40$ Volt
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Question 24 Marks
Draw a graph showing the change in potential corresponding to the change in distance from a point charge. For the charged system shown in the figure, prove that the potential difference between points $A$ and $B$ is
Image
Answer
In the figure, the potential at point A due to $+q$ and $-q$ charges respectively is
Image
$V _1=\frac{ l }{4 \pi \in_0} \frac{q}{a}$
And $\quad V _2=\frac{1}{4 \pi \in_0} \frac{(-q)}{(a+d)}$
Therefore the net potential at point A
$V_A=V_1+V_2$
$=\frac{1}{4 \pi \in_0}\left(\frac{q}{a}\right)+\frac{1}{4 \pi \in_0}\left(\frac{-q}{a+d}\right)$
$V _{ A }=\frac{1}{4 \pi \in_0}\left(\frac{1}{a}-\frac{1}{a+d}\right)$
$V _{ A }=\frac{q}{4 \pi \in_0}\left(\frac{a+d-a}{a(a+d)}\right)=\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}$
Similarly net potential at point B
$V _{ B }=\frac{q}{4 \pi \in_0}\left(\frac{1}{a+d}-\frac{1}{a}\right)$
$=\frac{-1}{4 \pi \in_0} \frac{q d}{a(a+d)}$
Therefore,
$V_A-V_B$$\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}-\left(\frac{-1}{4 \pi \in_0} \frac{q d}{a(a+d)}\right)$
$\begin{aligned} V _{ A }- V _{ B } & =\frac{1}{4 \pi \in_0} \frac{q d}{a(a+d)}[1+1] \\ & =\frac{2 q d}{4 \pi \in_0 a(a+d)}\end{aligned}$
$V _{ A }- V _{ B }=\frac{q}{2 \pi \in_0 a} \cdot \frac{d}{a+d}$


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