A constant electric field of magnitude 1 μNC^-1 exists along +x direction. What is the potential difference between points (1,0,0) and origin?

Question

Question

Answer 1

CONCEPT:

Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.

\(⇒ V=\frac{W}{q}\)

Potential due to a single charged particle Q at a distance r from it is given by:

\(⇒ V=\frac{Q}{4\piϵ_{0}r}\)

Where, ϵ0 is the permittivity of free space and has a value of 8.85 × 10-12 F/m in SI units

An Equipotential surface is a surface with a constant value of the potential at all points on the surface.
The following figures show equipotential surfaces due to a point charge and an electric dipole.

F16 Prabhu 27-4-2021 Swati D1 F16 Prabhu 27-4-2021 Swati D2

\(\Rightarrow E=-\frac{dV}{dx}\)

Where,
- E is the electric field
- V is the electric potential
- x is the direction of the electric field in space

EXPLANATION:

Given - E = 10^-6 NC^-1

\(\Rightarrow E=-\frac{dV}{dx}\)

\(\Rightarrow dV=-\int E.dx\)

\(\Rightarrow V_2-V_1=-\int_0 ^1 E.dx\)

\(\Rightarrow V_2-V_1=-E\int_0 ^1dx = -1 \mu \int_0 ^1dx = -10^{-6}\int_0 ^1dx = -10^{-6}x|_0^1= -10^{-6}(1-0)= -10^{-6} V\)

Additional Information

Potential at a point P due to a system of charged particles Q1, Q2, Q3, ... Qn having distances r1, r2, r3, ... rn respectively from point P is given by:

\(⇒ V=\frac{Q_1}{4\piϵ_{0}r} + \frac{Q_2}{4\piϵ_{0}r} + \frac{Q_3}{4\piϵ_{0}r} + ...+\frac{Q_n}{4\piϵ_{0}r}\)

\(⇒ V=\sum_{i=1}^{n}\frac{Q_i}{4\piϵ_{0}r_i}\)