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
Answers
Answered by
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
Answered by
1
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".
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