Question

A charge of q = - 4.0 × 10-6 is placed in an electric field and experiences a force of 5.5 N [E]

a) What is the magnitude and direction of the electric field at the point where charge q is located?

b) If charge q is removed, what is the magnitude and direction of the force exerted on a charge of - 2q at the same location as charge q?​

Answer 1

Answer:

a) The force on a charge q due to an electric field E is given by

F = q E

The magnitude of E is given by

| E | = | F | / | q | = 5.5 / (4.0 × 10-6) = 1.375 × 106 N / C

Since charge q is negative F and E have opposite direction. E is 1.375 × 106 N / C [W]

b) The force on a charge -2q due to an electric field E is given by

F2 = -2 q E = -2(q E) = -2(5.5[E]) = -11 [E] or 11 N [W]

Explanation:

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