A 10.0 g block with a charge of +8.00×10-5C is placed in an electric field E = (3000iˆ - 600 ˆj)N /Cs. What are the
(a) magnitude and
(b) direction (relative to the positive direction of the x-axis) of the electrostatic force on the block? If the block is released from rest at the origin at time t = 0, what
are its (c) x and (d) y coordinates at t = 3.00 s?

a) 24.5*10⁻² N b) θ = -11.3º c) x=108 m d) y=-21.6 m

Explanation:

a) Assuming that the block can be treated as a point charge, the electrostatic force on it, must be equal to the product of the electric field, times the value of  the charge.

At the same time, this force must obey Newton's 2nd law, as follows:

F = m*a = q*E

As this is a vector equation, and  we have the value of the x and y components of the electric field, we can decompose it in two algebraic equations, as follows:

Fₓ = m*aₓ = q*Eₓ

Fy = m*ay = q*Ey

In order to find the magnitude of the force F, we can find the magnitude of E, as follows:

The magnitude of the electrostatic force on the block is:

F = q*E = 8.00*10⁻⁵ C * 3.06*10³ N/C =24.5*10⁻² N

b) The direction of the force, as the charge is positive, by convention, is the same as the electric field.

The angle of the electric field with the positive x-axis, can be calculated from the values of Eₓ and Ey, as follows:

θ = tg⁻¹ (Ey/Ex) = tg⁻1 (-600/3000) = tg⁻1 (-.2) = -11.3º (11.3º below horizontal)

c) and d)

As the electric field is uniform, we can get the displacement due to the electrostatic force, applying kinematic equations, to the x and y directions, as they are independent each other due they are perpendicular each other.

As we have already told, we have the following algebraic equations:

Fₓ = m*aₓ = q* Eₓ

Fy = m*ay = q*Ey

As we have the values of Ex, Ey, and m, we can find aₓ and  ay, as follows:

We can find the horizontal and vertical displacements from the origin (the position coordinates at the time t), as follows:

⇒ x, y = 108.0 m, -21.6 m

it has to be b, about 97 percent.