Question

Which of the following figure shows the correct equipotential surfaces of'a system of two positive charges ?

A. Electric Potential mcq option image
B. Electric Potential mcq option image
C. Electric Potential mcq option image  
D. Electric Potential mcq option image
Answer :   Electric Potential mcq option image
Solution :
Electric Potential mcq solution image
Equipotential surfaces are normal to the electric field lines. The following figure shows the equipotential surfaces along with electric field lines for a system of two positive charges.

Releted MCQ Question on
Electrostatics and Magnetism >> Electric Potential

Releted Question 1

If potential (in volts) in a region is expressed as $$V\left( {x,y,z} \right) = 6xy - y + 2yz,$$      electric field (in $$N/C$$ ) at point $$\left( {1,1,0} \right)$$  is

A. $$ - \left( {3\hat i + 5\hat j + 3\hat k} \right)$$
B. $$ - \left( {6\hat i + 5\hat j + 2\hat k} \right)$$
C. $$ - \left( {2\hat i + 3\hat j + \hat k} \right)$$
D. $$ - \left( {6\hat i + 9\hat j + \hat k} \right)$$
Releted Question 2

A conducting sphere of radius $$R$$ is given a charge $$Q.$$ The electric potential and the electric field at the centre of the sphere respectively are

A. zero and $$\frac{Q}{{4\pi {\varepsilon _0}{R^2}}}$$
B. $$\frac{Q}{{4\pi {\varepsilon _0}R}}$$  and zero
C. $$\frac{Q}{{4\pi {\varepsilon _0}R}}{\text{and}}\frac{Q}{{4\pi {\varepsilon _0}{R^2}}}$$
D. Both and zero
Releted Question 3

In a region, the potential is represented by $$V\left( {x,y,z} \right) = 6x - 8xy - 8y + 6yz,$$       where $$V$$ is in volts and $$x,y,z$$  are in metres. The electric force experienced by a charge of $$2C$$ situated at point $$\left( {1,1,1} \right)$$  is

A. $$6\sqrt 5 N$$
B. $$30\,N$$
C. $$24\,N$$
D. $$4\sqrt {35} \,N$$
Releted Question 4

Four point charges $$ - Q, - q,2q$$   and $$2Q$$  are placed, one at each corner of the square. The relation between $$Q$$ and $$q$$ for which the potential at the centre of the square is zero, is

A. $$Q = - q$$
B. $$Q = - \frac{1}{q}$$
C. $$Q = q$$
D. $$Q = \frac{1}{q}$$

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