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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.

Ankita , 5 Years ago
Grade 12
anser 2 Answers
Pooja

Last Activity: 5 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

Last Activity: 5 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 = \sqrt{3^2+1^2+1^2} = \sqrt{11}, 
r^ = {(3-0)i + (1-0)j + (1-0)k} / |r|  =  (3i + j + k) / \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) / \sqrt{11}),
which is (K x 10 – 9 / 11\sqrt{11}) (3i + j + k).
 
Similarly, for charge Q, the point (3,1,1) is at diatance r = \sqrt{(3-2)^2+(1-0)^2+(1-0)^2} = \sqrt{3},
r^ = {(3-2)i + (1-0)j + (1-0)k} / \sqrt{3}  =  (i + j + k) / \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) / \sqrt{3}),
which is (K x Q/ 3\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\sqrt{11}) (3i)  +  (K x Q/ 3\sqrt{3}) (i) = K [ (3 x 10 – 9 / 11\sqrt{11}) + (Q/ 3\sqrt{3}) ] i
which is equal to zero.
\therefore  K [ (3 x 10 – 9 / 11\sqrt{11}) + (Q/ 3\sqrt{3}) ]  =  0  \Rightarrow  (3 x 10 – 9 / 11\sqrt{11}) + (Q/ 3\sqrt{3}) = 0
Solving, we get Q =  – (9\sqrt{3} / 11\sqrt{11}) x 10 – 9 C.

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