From a point charge, there is a fixed point $$A.$$ At $$A,$$ there is an electric field of $$500\,V/m$$ and potential difference of $$3000\,V.$$ Distance between point charge and $$A$$ will be
A.
$$6\,m$$
B.
$$12\,m$$
C.
$$16\,m$$
D.
$$24\,m$$
Answer :
$$6\,m$$
Solution :
$$\eqalign{
& E = 500\,V/m \cr
& V = 3000\,V. \cr} $$
We know that electric field $$\left( E \right) = 500 = \frac{V}{d}$$
$${\text{or}}\,\,d = \frac{{3000}}{{500}} = 6\,m$$
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