12.3.2 Electric Field of a Point Charge

Consider two point charges: the source charge Q and the test charge q, separated by distance r.

We can say that Q is exerting an electrical force of $\displaystyle {{F}_{e}}=\frac{{kQ}}{{{{r}^{2}}}}q$ on q, as according to Coulomb’s Law.

Alternatively, we can say that Q is creating an electrical field E around it, causing q to experience an electric force of qE.

Since electrical field strength is force per unit (positive) charge, the strength of the electric field produced by Q at distance r away is

$\displaystyle \begin{aligned}E&=\frac{{{{F}_{e}}}}{q}\\&=k\frac{{Qq}}{{{{r}^{2}}}}\div q\\&=k\frac{Q}{{{{r}^{2}}}}\end{aligned}$

Notice that the field strength obeys the inverse-square law.

What about the direction? Since a positive source charge will repel positive test charges radially outward, a positive source charge produces an electric field that is radially outward. Conversely, since a negative source charge will attract positive test charges radially inward, a negative source charge produces an electric field that is radially inward.

Example

Q_A and Q_B are point charges of +2.1 pC and −2.1 pC bound at vertices A and B of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate

a) the electric field strength at C.

b) the electric force experienced by
i) an electron
ii) a proton
placed at c

c) the electric field strength midpoint between A and B.

Solution

The magnitude of the field strength of Q_A at C,

$\displaystyle \begin{aligned}{{E}_{A}}&=k\frac{{{{Q}_{A}}}}{{{{r}^{2}}}}\\&=(8.99\times {{10}^{9}})\frac{{(2.1\times {{{10}}^{{-12}}})}}{{{{{(3.0\times {{{10}}^{{-6}}})}}^{2}}}}\\&=2.098\times {{10}^{9}}\text{ N }{{\text{C}}^{{-1}}}\end{aligned}$

The magnitude of the field strength of Q_B at C,

$\displaystyle \begin{aligned}{{E}_{B}}&=k\frac{{{{Q}_{B}}}}{{{{r}^{2}}}}\\&=(8.99\times {{10}^{9}})\frac{{(2.1\times {{{10}}^{{-12}}})}}{{{{{(3.0\times {{{10}}^{{-6}}})}}^{2}}}}\\&=2.098\times {{10}^{9}}\text{ N }{{\text{C}}^{{-1}}}\end{aligned}$

Note that Q_A exerts a radially outward field whereas Q_B exerts a radially inward field, E_A and E_B are thus in the directions shown in the diagram. The resultant field strength at C is thus given by the vector summation of E_A and E_B. Note that the vector summation triangle is an equilateral triangle. So the resultant field strength at C is

$\displaystyle {{E}_{R}}=2.098\times {{10}^{9}}\text{ N }{{\text{C}}^{{-1}}}\text{ (rightward)}$

$\displaystyle \displaystyle \begin{aligned}({{F}_{e}}&=qE)\\{{F}_{e}}&=(1.60\times {{10}^{{-19}}})(2.098\times {{10}^{9}})\\&=3.36\times {{10}^{{-10}}}\text{ N}\end{aligned}$

Since E_R is rightward, the proton experiences a rightward $\displaystyle \displaystyle 3.36\times {{10}^{{-10}}}\text{ N}$ force whereas the electron experiences a leftward $\displaystyle \displaystyle 3.36\times {{10}^{{-10}}}\text{ N}$ force.

At the midpoint, both E_A and E_B are rightward and have the same magnitude.

The resultant field strength is thus

$\displaystyle \begin{aligned}{{E}_{R}}&=2{{E}_{A}}&=2\times k\frac{Q}{{{{r}^{2}}}}&=2\times (8.99\times {{10}^{9}})\frac{{(2.1\times {{{10}}^{{-12}}})}}{{{{{(1.5\times {{{10}}^{{-6}}})}}^{2}}}}\\&=1.68\times {{10}^{{10}}}\text{ N }{{\text{C}}^{{-1}}}\end{aligned}$