Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the magnitude of electric field due to these charges at a point midway between them? (k = 1/4πε0 = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The electric field due to a point charge is given by:$$E = k\frac{q}{r^2}$$where:* k is Coulomb's constant (8.99 × 10^9 N ∙ m^2/C^2)* q is the charge (in Coulombs)* r is the distance from the charge (in meters)In this problem, we have two point charges:* q1 = +20.0 μC = +20.0 × 10^-6 C* q2 = -8.00 μC = -8.00 × 10^-6 CThe distance between the charges is:* r = 20.0 cm = 0.200 mThe point at which we want to find the electric field is midway between the charges, so the distance from each charge to the point is:* r1 = r2 = 0.100 mThe electric field due to q1 at the point is:$$E1 = k\frac{q1}{r1^2} = (8.99 × 10^9 N ∙ m^2/C^2)\frac{20.0 × 10^-6 C}{(0.100 m)^2} = 17.98 × 10^6 N/C$$The electric field due to q2 at the point is:$$E2 = k\frac{q2}{r2^2} = (8.99 × 10^9 N ∙ m^2/C^2)\frac{-8.00 × 10^-6 C}{(0.100 m)^2} = -7.19 × 10^6 N/C$$The net electric field at the point is the vector sum of E1 and E2:$$E = E1 + E2 = 17.98 × 10^6 N/C - 7.19 × 10^6 N/C = 25.2 × 10^6 N/C$$The electric field is directed toward the negative charge, so the correct answer is A.

Two Point Charges of +20