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

Step-by-step 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 (q_(1))/((r)^(2))`

Where,

`q_(1)` = 30 nC

r = 0.3 m

The electric field on point a due to
`q_(2)` charge,

`$E_(2)=k (q_(2))/((r)^(2))$`

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 (q_(1))/((r)^(2))+k (q_(2))/(\left(r^(\prime)\right)^(2))`

Now, putting all the values, we get;

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

`E_(n e t)=8.99 * 10^(9)(333.33+1,125)`

`E_(n e t) \approx 13,110 (N)/(C)`

Hence, the answer will be 13000 N/C directed to
`q_(2)`