A 5.81 g block with a charge of -6.37e -6 C is placed on a 24.7 degree incline.

Question

A 5.81 g block with a charge of -6.37e -6 C is placed on a 24.7 degree incline. If an electric field is simultaneously applied parallel to the incline, what must be its magnitude and direction so that the block will not move?

Please explain in detail to get points

A 5.81 g block with a charge of -6.37e -6 C is placed on a 24.7 degree incline. If an electric field is simultaneously applied parallel to the incline, what must be its magnitude and direction so that the block will not move? Please explain in detail to get points

Explanation / Answer

Electric Force = inclined component of Gravitational Force

E x Q = m x g x Sin 24.7

E = 0.00581 x 9.8 x Sin24.7 / (6.37 x 10^-6)

= 3735.08 N/C.

As the inclined component of gravitational force acts in the downward direction along the inclination, so the electric force must act in the upward direction along the inclination which can only be possible for given negative charge of the block if electric field acts in downward direction parallel to the inclination.

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