Question

Two small insulating spheres with radius 3.50×10^−2m are separated by a large center-to-center distance of
0.510m. One sphere is negatively charged, with net charge −2.40μC, and the other sphere is positively charged, with net charge 3.90μC. The charge is uniformly distributed within the volume of each sphere.

(A) What is the magnitude E of the electric field midway between the spheres?

(B) What is the direction of the electric field midway between the spheres?

Electric Field:

Electric field due to a charge at a point in space is the force experienced by a unit test charge when placed at that point.

Electric field at a point, at a distance r from the charge q, is given by E=kq/r^2
k=9×10^9Nm^2C^2

Answer 1

Answer:

Part (i) the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (ii) the direction of the electric field is towards the negative charge

Explanation:

Given;

+q = 3.93.90μC, r = 3.50×10⁻²m

-q = −2.40μC, r = 3.50×10⁻²m

magnitude of the electric field is experienced, midway between the spheres at a distance r,  r = ¹/₂ × 0.51 = 0.255 m

Electric field due to point charge is given as;

$E = \frac{F}{q} = \frac{Kq^2}{qr^2} = \frac{kq}{r^2}$

K is coulomb's constant = 8.99 x 10⁹ Nm²/C²

The positive charge on positive x-axis and the negative charge is on negative x-axis.

part (a)

The electric field due to positive charge; +q = 3.93.90μC

$E_+ = \frac{kq}{r^2}\\\\E_+ = \frac{8.99 X10^9*3.9X0^{-6}}{0.255^2}\\\\E_+= 5.3919 X10^5\frac{N}{C}$

The electric field due to negative charge; -q = −2.40μC

$E_- = \frac{kq}{r^2}\\\\E_- = \frac{8.99 X10^9*2.4X0^{-6}}{0.255^2}\\\\E_-= 3.3181 X10^5\frac{N}{C}$

From superimposition theorem

The magnitude of the electric field is;

E = E₊ + E₋

E = (5.3919 × 10⁵ + 3.3181 × 10⁵) N/C

E = 8.71 × 10⁵ N/C

Therefore, the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (b)

The direction of the electric field is towards the negative charge.