# A charge 10-9 C is located at the origin in free space and another charge Q at (2,0,0) . If the X component of the electric field at (3,1,1) is zero, calculate the value of Q.

Pooja
127 Points
4 years ago

We calculate field at (3,1,1) due to both charges.
E=E1+E2
=(kQ ÷ |r1|2)* vector component +
(kq ÷ |r2|2)* vector component
=kQ/3 *(i+j+k) +k×10-9/11 *(3i+j+k)
Now x component is zero
So , 0 =kQ/3 +3k×10-9/11
Q/3 = -3×10-9/11
So Q= -3/11 × 10-9 C

Samyak Jain
333 Points
4 years ago
Electric field due to a point charge Q along a unit vector r^ at a distance r from it is
(KQ / r2) r^.

For charge at origin, the point (3,1,1) is at distance r = $\dpi{80} \sqrt{3^2+1^2+1^2}$ = $\dpi{80} \sqrt{11}$
r^ = {(3-0)i + (1-0)j + (1-0)k} / |r|  =  (3i + j + k) / $\dpi{80} \sqrt{11}$
Electric field due to charge 10 – 9 C [located at origin (0,0,0)] at (3,1,1)  is  (K x 10 – 9 / 11)((3i + j + k) / $\dpi{80} \sqrt{11}$),
which is (K x 10 – 9 / 11$\dpi{80} \sqrt{11}$) (3i + j + k).

Similarly, for charge Q, the point (3,1,1) is at diatance r = $\dpi{80} \sqrt{(3-2)^2+(1-0)^2+(1-0)^2}$ = $\dpi{80} \sqrt{3}$,
r^ = {(3-2)i + (1-0)j + (1-0)k} / $\dpi{80} \sqrt{3}$  =  (i + j + k) / $\dpi{80} \sqrt{3}$.
Electric field due to charge Q [located at (2,0,0)] at (3,1,1)  is  (K x Q/ 3)((i + j + k) / $\dpi{80} \sqrt{3}$),
which is (K x Q/ 3$\dpi{80} \sqrt{3}$) (i + j + k).

x-component of electric field due to both the charges at (3,1,1) is
(K x 10 – 9 / 11$\dpi{80} \sqrt{11}$) (3i)  +  (K x Q/ 3$\dpi{80} \sqrt{3}$) (i) = K [ (3 x 10 – 9 / 11$\dpi{80} \sqrt{11}$) + (Q/ 3$\dpi{80} \sqrt{3}$) ] i
which is equal to zero.
$\dpi{100} \therefore$  K [ (3 x 10 – 9 / 11$\dpi{80} \sqrt{11}$) + (Q/ 3$\dpi{80} \sqrt{3}$) ]  =  0  $\dpi{100} \Rightarrow$  (3 x 10 – 9 / 11$\dpi{80} \sqrt{11}$) + (Q/ 3$\dpi{80} \sqrt{3}$) = 0
Solving, we get Q =  – (9$\dpi{80} \sqrt{3}$ / 11$\dpi{80} \sqrt{11}$) x 10 – 9 C.