Question

Two tiny particles having charges +60.0 mC and -12.0 mC are separated by a distance of 20.0 cm. What are the magnitude and direction of electric filed midwat between these two charges?

A. 64.7 × 10⁵ N/C directed towards the positive charge
B. 64.7 × 10⁶ N/C directed towards the positive charge
C. 64.7 × 10⁵ N/C directed towards the negative charge
D. 64.7 × 10⁶ N/C directed towards the negative charge
E. 64.7 × 10⁴ N/C directed towards the negative charge

Answer 1

Using Coulomb's Law and the formula for electric field, we can calculate the magnitude and direction of the electric field midway between two charges. The magnitude of the electric field is 16.2 x 10^9 N/C and it is directed towards the positive charge.

Step-by-step explanation:

The formula for electric field is given as:

Electric Field (E) = (k * Q) / r^2

where k is the electrostatic constant (9 x 10^9 N.m^2/C^2), Q is the charge on the particle, and r is the distance between the charges.

Plugging in the values in the formula, we have:

E = (9 x 10^9 N.m^2/C^2) * (60.0 mC - (-12.0 mC)) / (0.2 m)^2

E = (9 x 10^9 N.m^2/C^2) * (72.0 mC) / (0.04 m^2)

E = (9 x 10^9 N.m^2/C^2) * (0.072 C) / (0.04 m^2)

E = 16.2 x 10^9 N/C

The magnitude of the electric field midway between the two charges is 16.2 x 10^9 N/C.

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

Therefore, the correct answer is option A. 16.2 x 10^9 N/C directed towards the positive charge.