Question

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45nC. As shown in the diagram, point a is located 30 cm from q1 and 20 cm from q2. What are the magnitude and direction of the electric field at point a? Let the electrostatic constant k = 8.99 * 10 ^ 9 * N m ^ 2 / (C ^ 2)

A)7000 N/C, directed to q2
B)13,000 N/C, directed to q1
C)13,000 N/C, directed to q2
D)7000 N/C, directed to q1

Answer 1

The magnitude of the electric field at point a will be 13000 N/C towards $q_{2}$.

Answer: Option C

Explanation:

We know,

The distance between $q_{1}$ and point a = 0.3 m

The distance between $q_{2}$ and point a = 0.2 m

Now,

The electric field on point a due to $q_{1}$ charge,

                   $E_{1}=k \frac{q_{1}}{(r)^{2}}$

Where,

$q_{1}$ = 30 nC

r = 0.3 m

The electric field on point a due to $q_{2}$ charge,

$E_{2}=k \frac{q_{2}}{(r)^{2}}$

Where,

$q_{2}$ = 45 n C

r’ = 0.2 m

To find the total electric field exerted on point a due to both of the charges, we need to add $E_{1}$ and $E_{2}$.  Thus, the net electric field of point a,

$E_{n e t}=E_{1}+E_{2}$

$E_{n e t}=k \frac{q_{1}}{(r)^{2}}+k \frac{q_{2}}{\left(r^{\prime}\right)^{2}}$

Now, putting all the values, we get;

$E_{\text {net}}=k\left(\frac{30}{(0.3)^{2}}+\frac{45}{(0.2)^{2}}\right)$

$E_{n e t}=8.99 \times 10^{9}(333.33+1,125)$

$E_{n e t} \approx 13,110 \frac{N}{C}$

Hence, the answer will be 13000 N/C directed to $q_{2}$