A non-conducting solid sphere of radius R = 10.0 cm has a uniformly distributed charge Q = +1.50x10 C. Find the magnitude of the potential difference between a point at r = 50.0 cm and a point on the surface of the sphere. Select one:

Two point charges, A & B (qR = 123 µC), are separated by 86 cm. The distance between A & P is 20 cm. The potential at point P (see below) is zero. Determine qa. A 9A = 20 сm 86 cm WebAssign Plot

A small object with a mass of 270 μg carries a charge of 20.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 5.00 cm. If the thread makes an angle of 13.0° with the vertical, what is the potential difference between the plates?  V

Problem 13:   Each plate of a parallel-plate capacitor has an area of A = 0.61 m2. The plate separation is 3.0 mm, and one of these plates is shown in the figure. It carries a charge of Q = 2.9 μC, which is concentrated on its inner surface. Imagine a cylindrical Gaussian surface, of radius r = 0.012 m, whose axis is perpendicular to the plates. One end of the cylinder is inside the plate shown and the other end is located between the plates. Both ends are parallel to the plates. Part (a)  What is the flux through surface 1 Φ1, in newton meters squared per coulomb?  Part (b)  What is the flux through surface 2 (the outside of the cylinder not including the ends), in newton meters squared per coulomb?  Part (c)  What is the flux through surface 3 Φ3, in newton meters squared per coulomb?  Part (d)  Using all of your results, input an expression for the field within the capacitor E, in terms of the quantities given in the problem.

For a point charge, the equipotential surfaces form concentric spheres with the charges at the center. 2 such equipotential surfaces have potentials 293.3V and 95.2V. if these 2 concentric surfaces are separated by a distance of 2.2m, what is the charge of the point particle?

An air capacitor is made up of two parallel plates, each with an area of 10.7 cm^2 separated at a distance of 2.5 mm. A potential difference of 44 volts is applied to these plates. Calculate the surface charge density. answer nC/m^2 and two decimal places