Question

Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere is negatively charged, with net charge -2.50 μC, and the other sphere is positively charged, with net charge 4.20 μC. The charge is uniformly distributed within the volume of each sphere.

What is the magnitude E of the electric field midway between the spheres?
Take the permittivity of free space to be epsilon_0 = 8.85×10−12 C^2/(N \cdot m^2).
E =

\rm{N/C}

Answer 1

Answer:

$E = 7.83 \times 10^5 N/C$

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

$E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}$

here we know that

$q_1 = 4.20 \mu C$

$q_2 = 2.50 \mu C$

$r = \frac{0.555}{2}$

so we have

$E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}$

$E = 7.83 \times 10^5 N/C$

Answer 2

The  magnitude E of the electric field midway between the spheres is $E = 7.83 \times 10^5N/C$

Calculation of  magnitude E of the electric field:

When two-sphere should be oppositely charged due to this, net electric field at the mid point of two balls should be the sum of the electric field because of each ball at the mid point.

We know that

$E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2} \\\\= \frac{(9\times10^9)(4.20 + 2.50) \times 10^-6}{0.2775^2}$

$E = 7.83 \times 10^5N/C$

learn more about the sphere here: https://brainly.com/question/2888406