2 Marks Questions

Question 512 Marks

Define equipotential surface. Write its characteristics.

Answer

Definition : The surface in an electric field whose potential at every point is equal is called equipotential surface.Characteristics:
(i) The potential difference between any two points of it is zero.
(ii) No work has to be done in moving charge between any two of its points.
(iii) The electric field is always perpendicular to the equipotential surface.

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Question 522 Marks

Explain the dependence of the electric potential generated at a great distance due to a small electric dipole, when the point of observation is (i) in the axial position, and (ii) in the equatorial position.

Answer

(i) Potential in the axial position $ (V)=\frac{1}{4 \pi \epsilon_0}\left(\frac{p}{r^2}\right) $
Therefore it's clear $\quad V \propto \frac{1}{r^2}$
(ii) In equatorial position $V=0$, so it does not depend on distance. That is, the electric potential at each point in this situation is zero.

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Question 532 Marks

A point charge ' $q$ ' is placed at point $O$ as shown in the figure. Will $\left(V_A-V_B\right)$ be positive, negative or zero if $q$ is (i) positive (ii) negative.

Answer

Ans. $\quad V _{ A }- V _{ B }=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{O A}\right)-\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{O B}\right)$
$=\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{ OA }-\frac{1}{ OB }\right)$
$\because OA < OB , \therefore$ quantity $\left[\frac{1}{ OA }-\frac{1}{ OB }\right]$ will always be positive, therefore the sign of $\left(V_A-V_B\right)$ will depend on the sign of $q$.
$\left(\because 4 \pi \epsilon_0=\right.$ constant $)$
(1) If $q$ is positive then $\left(V_A-V_B\right)$ will be positive.
(2) When $q$ is negative, $\left( V _{ A }- V _{ B }\right)$ will be negative.

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Question 542 Marks

Define equipotential surface.

Answer

A surface on which the potential at every point is equal is called an equipotential surface. The value of potential difference between any two points on an equipotential surface is zero. Therefore, the work done in bringing a charge from one point to another point on an equipotential surface is equal to zero. But the work done is zero only when the charge is moved in a direction perpendicular to the electric field. Therefore, the equipotential surface is perpendicular to the electric field at every point.

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Question 552 Marks

Find the equivalence between A and B in the given circuit diagram.

Answer

Sol. Due to the equivalence at A being in series
$\frac{1}{ C }=\frac{1}{ C _1}+\frac{1}{ C _2}$
$\frac{1}{C}=\frac{1}{2}+\frac{1}{2}=1 \quad \therefore C =1 \mu F$
Similarly at $B$, from $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{ C }=\frac{1}{2}+\frac{1}{2}=1=1 \mu F$
$C =1 \mu F$
Therefore, capacitance between A and $B =1+1=2 \mu F$.

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Question 562 Marks

Three capacitors are arranged as shown in the circuit diagram. If the total capacitance between points A and B is $3 \mu F$, then the value of capacitance of X capacitor will be.

Answer

$\frac{1}{ C }=\frac{1}{3}+\frac{1}{x}=\frac{x+3}{3 x}$
$C =\frac{3 x}{x+3}$
$\therefore \quad \frac{3 x}{x+3}+1=3$
$\therefore \frac{3 x}{x+3}=2$
$x=6 \mu F$ Ans.

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Question 572 Marks

What will be the radius of a conducting sphere having electric capacitance of one Farad?

Answer

Given : $\quad C =1$ Farad , $R =$ ?
$\because \quad C =4 \pi \epsilon_0 R$
$\therefore \quad R =\frac{ C }{4 \pi \epsilon_0}=\frac{1}{4 \pi \epsilon_0} \times C$
$R =9 \times 10^9 \times 1=9 \times 10^9$ meter

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Question 582 Marks

Equipotential surfaces are 5 cm apart from each other. How much work will be required to move a 500 µC charge between distant points?

Answer

Potential difference between any two points on the post-isopotential surface $\Delta V =$ zero
Therefore work $W =$ charge $q \times$ potential difference $(\Delta V )$$=100 \mu C \times 0=\text { zero }$
i.e, no work has to be done.

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Physics 2 Marks Questions