Download electric potential/eqipotential

1) A metal sphere carries a charge 5*10-9 C and is at potential 0f 400V, relative to the
potential far away. the potential at the center of the sphere is:
For a charged sphere of radius r, If the charge is symmetrically distributed, the potential
outside a sphere of charge Q is given by the equation,
V = kQ/r
Where Q is the charge on the surface of the sphere.
r radius of the sphere
k is the coulomb’s constat = 8.89 * 10 -9 N m2 /c2
(k = 1/ 4 ∏ ε0 ; ε0 = 8.85 * 10 -12 F/ m
Since all the charge will reside on the conducting surface, according to Gaussian law,
there won’t be any charge inside the sphere.. So you see that if you substitute 0 for Q, the
potential at the center of the sphere is 0
2) A 5cm radius isolated conducting sphere is charged so its potential is +100V relative
to the potential far away. The charge density on the surface is:
Again using the same eqation of problem 1,
V = k Q/r
Q = V * r/K
Q = 100 * 5 * 10 – 2 / 8.89 * 10 -9 = 5.6 * 10 8 coulombs
I have used 5 * 10 – 2
for SI units.
because the distance is given in cm and all the units hold good
Charge density = Q/ Area of the sphere = 5.6 * 10 -8 / 4 ∏ r 2 = 1.7 * 10 6 coulombs/m2
[are of the sphere = 4 ∏ r 2 ]
3) A conducting sphere has charge Q and its electric potential is V, relative to the
potential far away. the charge is doubled to 2Q, the potential is:
V = k Q/r
Observe the V and Q are directly proportional
Now the charge is doubled, that means Q is now 2Q
So, V = k2Q/r
Potential is also doubled.
4) The potential difference between its ends of a 2m stick that is parallel to a uniorm
electric field is 400V. The magnitude of the electric field is:
Potential and electric field are related by
E=V/r
V = 400V r = 2
E = 400/ 2 = 200 V /m
5) The work required to carry a particle with charge of 6C from 5V eqipotential surface
to a 6V eqipotential surface and back again to the 5V surface is:
Work done is given by the formula W = Q (∆ V)
∆ V = V1 – V2 = 6 – 5 = 1 V
W = 6 * 1 = 6 Coulombs – Volt
Since the charge is again brought back to it’s original spot, there is the same amount of
work done again,
So total work W = 6 + 6 = 12 coulombs- Volts