Answer :


Given:
Magnitude of Electric field: E = 20 N C–1
E is along x-axis
Formula used:
As Electric field is along x-axis, potential difference will be along x-direction. Which means only x co-ordinates will be considered.
We know that,
Here dV is the change in potential : dV= VB-VA
E is the electric field along positive x axis and ds is the change in displacement.
(a)
A = (0, 0); B = (4m, 2m)
VB-VA= -20×(4-0) = -80 V
(b)
A = (4m, 2m); B = (6m, 5m)
VB-VA= -20×(6-4) = -40 V
(c)
A = (0, 0); B = (6m, 5m)
VB-VA= -20×(6-0) = -120 V
(d)
From (a),(b) and (c), we conclude that:
Potential difference of at points A = (0, 0), B = (6m, 5m)
= Potential difference at points A = (0, 0), B = (4m, 2m)
+ Potential difference at points A = (4m, 2m), B = (6m, 5m)


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

(a) In a quark moPhysics - Exemplar

A uniform field oHC Verma - Concepts of Physics Part 2

The work done to Physics - Exemplar

A uniform electriHC Verma - Concepts of Physics Part 2

Equipotential surPhysics - Exemplar

If a conductor haPhysics - Exemplar

Two charges q<subPhysics - Exemplar

Two charges –q eaPhysics - Exemplar

Two point chargesPhysics - Exemplar

Calculate potentiPhysics - Exemplar