Question

The magnitude of point charge due to which the electric field 30 cm away has the magnitude 2 NC-1 will be (A) 11 2 10 C u (B) 11 3 10 C u (C) 11 5 10 C u (D) 11 9 10 C

Answer 1

Answer:

Explanation:

Distance (r) = 0.3m

Electric Field (E) =  2 NC^-1

Charge (Q) = ?

Constant(k) = 8.99 x 10^9 N m^2/C^2

E = k Q /r2

2 = (8.99 x 10^9 * Q)/(0.3 * 2)

2    = 9 *109 *Q /(.3)2

Q = 2 * 10 -11  C

Answer 2

Answer:

Charge Q = "2.002*(10^-11) Cu".

Explanation:

Solution:

Given data

Distance =  R = 3 cm = 0.3 m

Electric field = E = 2 NC^-1

Required :

charge = Q= ?

Calculation:

We known that

E= K Q / (R^2)   ...................(1)

where K is the constant its numerical value is "8.99 x 10^9 N m^2/C^2"

             Putting values in equation (1) we get

2 = 8.99 x 10^9 *(Q) /( (0.3)^2)

Q = 2 * (0.09)  / 8.99 x 10^9

Q = 0.02002*(10^-9) Cu

Q = 2.002*(10^-11) Cu.

SO charge is equal to "2.002*(10^-11) Cu".