Question

A charge of q = - 6.0 × 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 - 3q at the same location as charge q? ​

Answer 1

Answer:

The magnitude of the electric field at the point where charge q is located can be determined using the equation =  -916.7 N/C

Explanation:

From the above question,

They have given :

A charge of q = - 6.0 × 10-6 is placed in an electric field and experiences a force of 5.5 N [E]

a) The magnitude of the electric field at the point where charge q is located can be determined using the equation:

                               E = F/q

where F is the force on the charge and q is the charge.

Plugging in the values given, we get:

                              E = 5.5 N/ (-6.0 x 10-6 C)

                                 = -916.7 N/C

The direction of the electric field is in the opposite direction of the force (negative).

b) The magnitude of the force exerted on a charge of -3q at the same location as charge q can be determined using the equation:

                              F = qE

where q is the charge and E is the electric field.

Plugging in the values given, we get:

                              F = (-3)(-916.7 N/C)

                                 = 2750 N

                              F = 16.5 N

The direction of the force is in the opposite direction of the force (negative).

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