In bringing an electron towards another electron, the electrostatic potential energy of the system
A.
decreases
B.
increases
C.
remains same
D.
becomes zero
Answer :
increases
Solution :
The electron has negative charge. When an electron is brought towards another electron, then due to same negative charges repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases. Alternative
Electrostatic potential energy of system of two electrons $$U = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { - e} \right)\left( { - e} \right)}}{r}$$
$$ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{r}$$
Thus, as $$r$$ decreases, potential energy $$U$$ increases.
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.
$$ - \left( {6\hat i + 9\hat j + \hat k} \right)$$
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.
$$\frac{Q}{{4\pi {\varepsilon _0}R}}{\text{and}}\frac{Q}{{4\pi {\varepsilon _0}{R^2}}}$$
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
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