Question

6.At what distance from a positive charge of 8.421 mC would the electric field strength be 3.91 x 103 N/C​

Answer 1

Explanation:

that was an easy one.

go with the solution you will better understand it

Answer 2

Concept:

• Electric field strength
• An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity.
• Additionally, it refers to a system of charged particles' physical fields.

Given:

• Charge q = 8.421 mC = 8.421 * 10^-3 C
• Electric field strength E = 3.91 x 10^3 N/C​

Find:

• The distance from the positive charge of 8.421 * 10^-3 C where the electric field strength is 3.91 x 10^3 N/C​

Solution:

We know that the electric field is

E = kQ/r^2

E = K * 8.421 * 10^-3 C /r^2

r^2 = K * 8.421 * 10^-3 C/ E

r^2 = K * 8.421 * 10^-3 C/ 3.91 x 10^3 N/C

K = 9*10^9

r^2 = 9*10^9 * 8.421 * 10^-3 C/ 3.91 x 10^3 N/C

r^2 = 19.38 * 10^3

r = 1.39 m

The distance is 1.39 m

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