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2 years ago

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Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3

cm by 2 cm) the magnitude of the electric field is 400 V/m, and on the right face the magnitude of the electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux, and the total amount of charge that is inside the box.

1 answer:

natima [27]2 years ago

7 0

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

<h3>Area of the left face</h3>

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

<h3>Electric flux on the left face</h3>

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

<h3>Area of the right face</h3>

A2 = 0.05 m x 0.02 m = 0.001 m²

<h3>Electric flux on the right face</h3>

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

<h3>Total flux</h3>

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

<h3>Total charge inside the box</h3>

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

Learn more about total flux here: brainly.com/question/26289097

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