# A large conductin

Given,

Surface charge density,σ1.0 × 10–4 C m–2

Edge length of the cube, e=1.0 cm=0.01m

Permittivity of free space, ε0= 8.85×10-12 F/m

Formula used

For a conducting plate infinite length), the electric field, E is,

And the electrostatic energy density or the energy per volume is,

Substituting eqn.1 in eqn.2 will result in,

Now the energy stored in volume V is

In the problem, we have to find the force inside a cube of edge e length.

So, we replace V with e3 in eqn.3

So,

Now, substituting the known values in the above equation, it becomes,

Or,

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
view all courses
RELATED QUESTIONS :

A capacitor is maPhysics - Exemplar

A parallel plate Physics - Exemplar

The battery remaiPhysics - Exemplar

A parallel-plate HC Verma - Concepts of Physics Part 2

How many time conHC Verma - Concepts of Physics Part 2

The plates of a cHC Verma - Concepts of Physics Part 2

A capacitor of caHC Verma - Concepts of Physics Part 2