Question

5. Compute the electric field experienced by a test charge q = + 0.80 µC from a source charge q = + 15 µC in a vacuum when the test charge is placed 0.20 m away from the other charge. (Calculate the electrostatic force and electric field.)

Answer 1

Explanation:

In vector calculus notation, the electric field is given by the negative of the gradient of the electric potential, E = −grad V. This expression specifies how the electric field is calculated at a given point. Since the field is a vector, it has both a direction and magnitude.

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Answer 2

Answer: F= 168.75 N/C

Explanation:

Formula:

F = k * (Q / r^2)

Given:
F = ?
k = 9x10^9 N m^2/C^2
q = 15 x 10^-6 C
r = 0.20m

Solution:
F = (9 x 10^9 N m^2/C^2) * (15 x 10^-6 C) / (0.20 m)^2

F = 168.75 N/C