What is the magnitude of the electric field at a point 0.0050 m from a 0.0040

C charge?

Use E = kg and k = 9.00 x 10°N. m/c?

A. 1.4 x 1072 N

B. 2.8 1012 N

C. 1.1 x 1010 N

D. 7.2×10°N

A. 1.4 * 10^(12) N/C

Explanation:

Parameters given:

Distance of charge from electric field, r = 0.0050 m

Charge, q = 0.0040 C

The electric field at a point, r, away from a charge, q, is given as:

E = (k * q) / r²

Where k = Coulomb's constant

E = (9 * 10^9 * 0.004) / (0.005²)

E = 1440 * 10^9 = 1.4 * 10^(12) N/C

The magnitude of the electric field at a point 0.005 m from the charge is 1.4 * 10^(12) N/C.

Explanation:

Given that,

A charge Q = 0.004 C

We want to find the electric field at distance r = 0.005m

K = 9 × 10^9 Nm²/C²

Electric field at a point is given as

E = kQ / r²

E = 9 × 10^9 × 0.004 / 0.005²

E = 1.44 × 10¹² N/C

So, the magnitude of the electric field at 0.005m is 1.44 × 10¹² N/C.

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.