Question

A +29 nc charge is at the origin. how much farther from the charge is the 2000 v equipotential surface than the 3000 v surface?

Answer 1

The general formula for calculating charge is given by:
V=kq/r where V=Equipotential, k= Coulomb's constant = 9*10^9 Nm^2C^-2 and q= Charge in coulombs, and r= distance from to the equipotentiality surface.

Charge = +29 nc =29*10^-9 C

At 1000 V, r= kq/V = (9*10^9*29*10^-9)/1000 = 0.261 m
At 3000 V, r=kq/V = (9*10^9*29*10^-9)/3000 = 0.087 m

1000 V surface is (2.61-0.87) m further than 3000 V surface from the charge.

That is, 0.261-0.087 = 0.174 m

Answer 2

Answer:

0.0436m

Explanation:

The potential (V) of a point charge (Q) at a distance (r) is given by;

V = k x Q / r         ------------------(i)

Where;

k = constant =  9 x 10⁹ Nm²/C²

[A] Now,

(i) let's get the distance (r₁) from the charge with the 2000V equipotential as follows;

2000 = k Q / r₁

Make r₁ subject of the formula as follows;

r₁ = k Q / 2000

Therefore the distance r₁ of the 2000V equipotential surface from the charge is  kQ / 2000

(ii) let's get the distance (r₂) from the charge with the 3000V equipotential as follows;

3000 = k Q / r₂

Make r₂ subject of the formula as follows;

r₂ = k Q / 3000

Therefore the distance r₂ of the 3000V equipotential surface from the charge is  kQ / 3000

[B] To get how much farther (r) from the charge the 2000V equipotential surface is than the 3000V, we find the difference between r₁ and r₂ as follows;

r = r₁ - r₂         ------------------(ii)

Where;

r₁ = k Q / 2000

r₂ = k Q / 3000

Substitute these values into equation (ii) as follows;

r = $\frac{k Q}{2000}$ - $\frac{k Q}{3000}$

Solve for r;

r = 0.0005kQ - 0.0003kQ

r = kQ (0.0005 - 0.000333)

r = kQ (0.000167)

Now substitute the values of k and Q = 29n C (29 x 10⁻⁹C) into the equation above as follows;

r = 9 x 10⁹ x 29 x 10⁻⁹ x 0.000167

r = 0.0436m

Therefore, the 2000V equipotential surface is 0.0436m from the 3000V surface.