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Questions OBJECTIVE - I and Answer Electric Field and Potential HC Verma Part II


Figure (29-Q2) shows some of the electric field lines corresponding to an electric field. The figure suggests that

(a) $E_A >E_B>E_C$ 

(b) $E_A =E_B=E_C$  

(c) $E_A =E_C>E_B$  

(d) $E_A =E_C<E_B$  

Answer: (c).  

There are an equal number of electric field lines within the same areas at A and C but fewer lines at B. 

When the separation between two charges is increased the electric potential energy of the charges

(a) increases

(b) decreases

(c) remains the same

(d) may increase or decrease   

Answer: (d).  

Electric potential energy is defined as negative of the work done by the electric forces. If both the charges are of the same nature the work done by the forces is positive and potential energy is negative on increasing the separation. But it will be the opposite if the charges are of a different nature.



If a positive charge is shifted from a low potential region to a high potential region, the electric potential energy

(a) increases

(b) decreases

(c) remains the same

(d) may increase or decrease.    

Answer: (a). 

In shifting a positive charge from low potential energy region to high potential energy region work will be done against the electric force, i.e. the work done by the electric force is negative so the potential energy is positive, i.e. the electric potential energy increases. 


Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential

(a) continuously increases

(b) continuously decreases

(c) increases then decrease

(d) decreases then increase.    

Answer: (d).  

Suppose two positive charges q and q are placed at distnce r from each other. Potential at a point P at a diatance x from A (x<r) due to these charges will be

$V= \frac{q}{4\pi \epsilon_0 x}+\frac{q}{4\pi \epsilon_0 (r-x)}$

$V= \frac{q}{4\pi \epsilon_0}\left(\frac{r-x+x}{x(r-x)}\right)$

$V= \frac{qr}{4\pi \epsilon_0x(r-x)}$

differentiating V with respect to x we get

$\frac{dV}{dx}=\left(\frac{qr}{4\pi \epsilon_0}\right) \left[-\frac{1}{x^2(r-x)}+\frac{1}{x(r-x)^2)}\right]$

$\frac{dV}{dx}=\left(\frac{qr}{4\pi \epsilon_0x(r-x)}\right) \left[\frac{1}{(r-x)}+\frac{1}{x}\right]$

$\frac{dV}{dx}=\frac{qr(2x-r)}{4\pi \epsilon_0x^2(r-x)^2}$

At x → 0, $\frac{dV}{dx}$→Infinity(-ve)

at x = $\frac{r}{2}$, $\frac{dV}{dx}=0$

at x →r, $\frac{dV}{dx}$ →Infinity(+ve)

So the potential first decreases and after x = $\frac{r}{2}$ it increases.   


The electric field at the origin is along the positive X-axis. A small circle is drawn with the center at the origin cutting the axes at points A, B, C and D having coordinates (a,0), (0,a), (-a,0), (0,-a) respectively. Out of the points on the periphery of the circle, the potential is minimum at

(a) A

(b) B

(c) C

(d) D

Answer: (a).  

Since the potential decreases at the maximum rate along the electric field, the minimum potential among these four will be minimum at the point on the x-axis i.e. at A.


If a body is charged by rubbing it, its weight

(a) remains precisely constant

(b) increases slightly

(c) decreases slightly

(d) may increase slightly or may decrease slightly.     

Answer: (d).  

When a body is charged by rubbing, either it gets some electrons from another body to be negatively charged or loses some electrons to be positively charged. So either its weight may increase slightly or decrease slightly due to the elctron transfers.


An electric dipole is placed in a uniform electric field. The net electric force on the dipole 

(a) is always zero

(b) depends on the orientation of the dipole

(c) can never be zero

(d) depends on the strength of the dipole.   

Answer: (a).  

In a uniform electric field both ends will experience equal and opposite force (because the ends of a diapole have equal and opposite charge) hence the net force on the diapole will always be zero. (But there will be the torque).


Consider the situation of the figure (29-Q3). The work done in taking a point charge from P to A is WA, from P to B is WB and from P to C is WC.

(a) $W_A < W_B<W_C$ 

(b) $W_A > W_B>W_C$

(c) $W_A = W_B=W_C$

(d) None of these   

Answer: (c)

The three points A, B and C are at equal distances from q. So the potential at each of the three points will be equal. Thus the work done in bringing a point charge from P to these points will be equal. 


A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is

(a) zero

(b) positive

(c) negative

(d) zero if the charge Q is at the center and nonzero otherwise.   

Answer: (a).  

The work done by the electric field depends upon the final and initial positions, not on the path followed by the charge. In one complete revolution, the rotating charge comes to the same point hence no work is done.  


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