The figure shows a box on whose surfaces the electric field is measured to be ho

Question

The figure 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 E1 is 320 volts/meter, and on the right face the magnitude of the electric field E2 is 1090 volts/meter. 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.

a). Flux through left : ? V/M

b). Flux through the right: ? V/M

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Explanation / Answer

a) The flux is simply a product of electric field and area. Area = 0.02m * 0.03m = 6 * 10^(-4)m^2. Therefore flux is 320 * 6 * 10^(-4) = 0.192Vm

b) Similarly, 1090 * 6 * 10^-4 = 0.654Vm

On the other sides, the electric flux is 0, because the electric field normal to the surface is zero(even though there are tangential components.)
Now we can use Gauss' law. Be careful because on the left the flux is going in, while on the right the flux is going out. Hence the total electric flux coming out of the box is
0.654Vm - 0.192Vm = 0.462 Vm = Q/

Q= 8.85 * 10^(-12) * 0.462 = 4.09 * 10^(-12)C(Coulomb)

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