The electric flux through a surface is given by the dot product of the electric field vector and the area vector. The area vector is perpendicular to the surface and its magnitude is equal to the area of the surface.

In this case, the circle is in the xz-plane, so its area vector is in the y-direction. The electric field has components in the x, y, and z directions, but only the y-component contributes to the flux because the dot product of the electric field vector and the area vector involves the cosine of the angle between them, which is 0 for the x and z components (cosine of 90 degrees is 0) and 1 for the y component (cosine of 0 degrees is 1).

The magnitude of the electric field in the y-direction is 1.30×10^3 N/C. The area of the circle is πr^2, where r is the radius. The diameter is given as 15.0 cm, so the radius is 7.5 cm or 0.075 m. Therefore, the area is π*(0.075 m)^2 = 0.0177 m^2.

The electric flux is then (1.30×10^3 N/C) * (0.0177 m^2) = 23.01 N*m^2/C.

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The electric flux through a surface is given by the dot product of the electric field vector and the area vector. The area vector is perpendicular to the surface and its magnitude is equal to the area of the surface.

In this case, the circle is in the xz-plane, so its area vector is in the y-direction. The electric field has components in the x, y, and z directions, but only the y-component contributes to the flux because the dot product of the electric field vector and the area vector involves the cosine of the angle between them, which is 0 for the x and z components (cosine of 90 degrees is 0) and 1 for the y component (cosine of 0 degrees is 1).

The magnitude of the electric field in the y-direction is 1.30×10^3 N/C. The area of the circle is πr^2, where r is the radius. The diameter is given as 15.0 cm, so the radius is 7.5 cm or 0.075 m. Therefore, the area is π*(0.075 m)^2 = 0.0177 m^2.

The electric flux is then (1.30×10^3 N/C) * (0.0177 m^2) = 23.01 N*m^2/C.

Knowee AI · Accepted Answer

The electric flux through a surface is given by the dot product of the electric field vector and the area vector. The area vector is perpendicular to the surface and its magnitude is equal to the area of the surface.

In this case, the circle is in the xz-plane, so its area vector is in the y-direction. The electric field has components in the x, y, and z directions, but only the y-component contributes to the flux because the dot product of the electric field vector and the area vector involves the cosine of the angle between them, which is 0 for the x and z components (cosine of 90 degrees is 0) and 1 for the y component (cosine of 0 degrees is 1).

The magnitude of the electric field in the y-direction is 1.30×10^3 N/C. The area of the circle is πr^2, where r is the radius. The diameter is given as 15.0 cm, so the radius is 7.5 cm or 0.075 m. Therefore, the area is π*(0.075 m)^2 = 0.0177 m^2.

The electric flux is then (1.30×10^3 N/C) * (0.0177 m^2) = 23.01 N*m^2/C.

The electric field in a region of space can be expressed by (245i^𝑖^ + 1.30×103j^𝑗^ + 851k^𝑘^) N/C. If a circle of diameter 15.0 cm is placed in the xz-plane of this region, what will be the electric flux through the circle?

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