Energy And Electricity Quiz

The electric field due to a charged sphere at a point outside the sphere can be calculated using the formula E = kQ/r^2, where E is the electric field, k is the electrostatic constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. Plugging in the given values, we get E = (9 × 10^9 N m^2/C^2)(3 × 10^-6 C)/(0.06 m)^2 = 7.5 × 10^6 N/C.

As the plates of a parallel-plate capacitor are pulled apart, the distance between them increases. The strength of the electric field between the plates is directly proportional to the voltage applied and inversely proportional to the distance between the plates. Therefore, as the distance increases, the strength of the electric field decreases.

The electric field at a distance r from the axis of an infinitely long cylinder with uniform charge density can be calculated using the formula E = (Sigma * r)/(2 * epsilon), where Sigma is the charge density, r is the distance from the axis, and epsilon is the permittivity of free space. Plugging in the given values, we get E = (18 μC/m3 * 1 cm)/(2 * epsilon). Simplifying this expression gives E = 9 μC/(2 * epsilon). Using the value of epsilon, which is approximately 8.85 x 10^-12 C2/(N*m2), we can calculate E to be approximately 10.2 x 103 N/C.

When a positive object touches a neutral electroscope, the leaves separate because the positive charges are repelled and move away from each other. However, when a negative object is brought near the electroscope, the negative charges attract the positive charges in the electroscope, causing them to move closer together. Therefore, the correct answer is "They move closer together."

The potential difference between two points in an electric field is given by the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the two points. In this case, the electric field is given as 12 V/m in the positive x-direction. The distance between the origin (0,0) and the point (3.0 m, 4.0 m) can be calculated using the Pythagorean theorem, which gives a distance of 5.0 m. Plugging these values into the formula, we get V = (12 V/m)(5.0 m) = 60 V. However, since the question asks for the potential difference with the origin at the higher potential, the answer is 36 V.

The electric field due to a line of charge is inversely proportional to the distance from the line. Since Lambda1 is greater than Lambda2, the electric field due to line #1 is stronger than the electric field due to line #2. Therefore, there must be a point between the lines where the electric fields due to line #1 and line #2 cancel each other out, resulting in zero net electric field. This point will be closer to line #2 than line #1. Hence, the locus of points where the electric field is zero is along a line between the lines closer to line #2 than line #1.

A negative charge, being attracted to positive charges, naturally moves from areas of low potential (where there are fewer positive charges) to areas of high potential (where there are more positive charges). This movement allows the negative charge to reduce the overall potential energy of the system and reach a more stable state. Therefore, a negative charge tries to move from low potential to high potential.

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. In this case, all wires are made of the same material, so the resistivity is constant.

Comparing the wires, we can see that the wire with the largest resistance will have the longest length and the smallest diameter.

Therefore, the wire of length L and diameter d/2 will have the largest resistance because it has the longest length and the smallest diameter among the given options.

The electric potential is zero at a point on the x-axis when the sum of the electric potential due to the positive charge QA and the electric potential due to the negative charge QB is equal to zero. In this case, the electric potential due to QA is positive and the electric potential due to QB is negative. Since QB has a greater magnitude, the electric potential due to QB will be larger. Therefore, the point where the electric potential is zero will be closer to QB. Since QB is located at x = d and the point x = d/4 is closer to QB than any other option, the electric potential is equal to zero at x = d/4.

When a dielectric is inserted between the plates of a parallel-plate capacitor, it increases the capacitance of the capacitor. As the potential energy stored in a capacitor is directly proportional to the capacitance, the potential energy decreases when the capacitance increases. Therefore, the potential energy of the capacitor decreases when a dielectric is inserted between the plates.

The peak current flowing through a device can be calculated using the formula I = P/V, where I is the current, P is the power, and V is the voltage. In this case, the power is 500 W and the voltage is 120 V. Plugging these values into the formula, we get I = 500/120 = 4.2 A. Therefore, the correct answer is 4.2 A.

Tilting the frame so that its plane is now in the yz-plane would decrease the magnitude of the electric flux through the frame. This is because the electric field is set up along the y-axis, and when the frame is tilted to the yz-plane, it will no longer be perpendicular to the electric field. The electric flux is maximized when the frame is perpendicular to the electric field, so any change that deviates from this orientation will decrease the magnitude of the electric flux.

The potential energy stored in a capacitor is given by the formula U = (1/2) * C * V^2, where U is the potential energy, C is the capacitance, and V is the voltage across the capacitor. In this case, the capacitance is given as 6.00 μF. The voltage can be calculated using the formula V = Q / C, where Q is the charge on the capacitor. Since the charge on each plate is +40.0 μC and -40.0 μC, the total charge is 80.0 μC. Plugging these values into the formula, we get V = 80.0 μC / 6.00 μF = 13.3 V. Finally, substituting the values of C and V into the formula for potential energy, we get U = (1/2) * 6.00 μF * (13.3 V)^2 = 133 μJ.