Key Idea : Search for the relations of electric potential and electric field at a particular point. At any point, electric potential due to charge Q is V=4πε01⋅rQ where r is the distance of observation point from the charge. At the same point, electric field is E=4πε01⋅r2Q Combining Eqs. (i) and (ii), we have E=Q4πε0V2=Q4πε0×(Q×1011)2=4πε0Q×1022 V/m
The electric potential at a point in free space due to a charge Q coulomb is Q × 10^11 ~V. The electric field at that point is
NEET UG 2008 — Physics Electromagnetism
The electric potential at a point in free space due to a charge Q coulomb is Q×1011 V. The electric field at that point is
Held on 30 Apr 2008 · Verified 9 Jul 2026.
Options
- A
4πε0Q×1022 V/m
- B
12πε0Q×1020 V/m
- C
4πε0Q×1020 V/m
- D
12πε0Q×1022 V/m
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
NEET UG Electromagnetism in other years
More NEET UG Electromagnetism questions
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)
A 2 amp current is flowing through two different small circular copper coils having radii ratio $1: 2$. The ratio of their respective magnetic moments will be
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $\mathrm{c} / 100(\mathrm{c}=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \mathrm{~T}$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Other Physics topics for NEET UG
Keep practicing Electromagnetism
Work through every NEET UG Electromagnetism PYQ, year by year.