Question

A block with a mass of 13.1 g and a charge of +5.38 × 10^-5 C is placed in an electric field with x component E_x = 2.32 × 10³ N/C, y component E_y = -856 N/C, and z component E_z = 0. (a) What is the magnitude of the electrostatic force on the block and (b) what angle does that force make with the positive x direction? 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 = 4.05 s?

(a)

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(b)

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(c)

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(d)

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Answer 1

The details given are as follows

Mass of the block , m = 13.1g = 13.1 x 10^-3 kg

Charge of the block , q = 5.38 x 10^-5 C

X component of the Electric Field , E_x = 2.32 x 10³ N/C

Y component of the Electric Field , E_y = -856 N/C

(a)

Note that both the Electric Field and Force are vectors, however we are concerned only about the magnitude of these quantities. Which are represented by,

|E| for Electrostatic Field

|F| for Electrostatic Force

|F| and |E| are related by the equation |F| = q|E|

|E| = √ ( E_x² + E_y² )

= √ ( (2.32x10³)² + (-856)² )

= √ ( 5.38x10⁶ + 7.32x10⁵)

= 2472.88 N/C

Therefore the Electrostatic Force ,

|F| = q |E|

= (5.38x10^-5 ) x ( 2472.88)

= 0.133N

(b)

The Force and Electric Field are in the same direction.

Therefore the angle the force makes with the X direction is the same as that made by the Electric Field in the X direction. Which is given by the formula

tan θ = Ey / E_x

= -856 / 2.32x10³

= -0.3689

θ = tan^-1( -0.3689)

= -20.25^o  

(c)

To find the distance travelled in the X direction or the X coordinate we must first find the force acting in the X direction, which is given by

F_x = q E_x

= ( 5.38x10^-5 ) * (2.32x10³)

= 0.1248N

But from Newtons Law F = ma

Therefore acceleration in the X direction can be found as follows

m a_x = q E_x

a_x = q E_x / m

= 0.1248 / 13.1x10^-3

= 9.52 m/s²

Using the formula S = ut + 1/2 at²

S is the distance travelled , t is the time of travel , u is the initial velocity , a is the acceleration

Here initial velocity , u = 0

t = 4.05 s

S = 1/2 at²

= 1/2 *(9.52)* (4.05)²

= 78.07m

(d)

To find the Y Coordinates we follow the same method used to find the X coordinate

Fy = q Ey

= ( 5.38x10^-5 ) * (-856)

= -0.046N

From Newtons Law

m a_y = q E_y

a_y = q E_y/ m

= -0.046 / 13.1x10^-3

= -3.51 m/s²

Using the formula S = ut + 1/2 at²

Here initial velocity , u = 0

t = 4.05 s

= 1/2 *(-3.51)* (4.05)²

= - 28.78m