You are watching: Find the potential at point a.

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge \(q\) from a large distance away to a distance of \(r\) from a point charge \(Q\), and noting the connection between work and potential \((W=-q\Delta V)\), we can define the *electric potential \(V\)** of a point charge:*

definition: ELECTRIC POTENTIAL \(V\) OF A POINT CHARGE

The electric potential \(V\) of a point charge is given by

\

where \(k\) is a constant equal to \(9.0 \times 10^{9}\, \mathrm{N}\cdot \mathrm{m^{2}/C^{2}}.\)

The potential at infinity is chosen to be zero. Thus \(V\) for a point charge decreases with distance, whereas \(\mathbf{E}\) for a point charge decreases with distance squared:

\

Recall that the electric potential \(V\) is a scalar and has no direction, whereas the electric field \(\mathbf{E}\) is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as *vectors*, taking magnitude and direction into account. This is consistent with the fact that \(V\) is closely associated with energy, a scalar, whereas \(\mathbf{E}\) is closely associated with force, a vector.

Example \(\PageIndex{2}\): What Is the Excess Charge on a Van de Graaff Generator

A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (Figure \(\PageIndex{1}\)) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

Figure \(\PageIndex{1}\): The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.**Strategy**

The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using Equation \ref{eq1}.

**Solution**

Solving for \(Q\) and entering known values gives

\< \begin{align*} Q &=\dfrac{rV}{k} \\<5pt> &= \dfrac{(0.125 \,\mathrm{m})(100\times 10^{3}\, \mathrm{V})}{8.99\times 10^{9}\, \mathrm{N\cdot m^{2}/C^{2}}} \\<5pt> &= 1.39\times 10^{-6} \,\mathrm{C} \\<5pt> &= 1.39\, \mathrm{\mu C}.\end{align*}\>

**Discussion**

This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.

See more: Which Depends On Location Weight Or Mass And Weight, What Is The Difference Between Weight And Mass