Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        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.
7 months ago

Pooja
127 Points
							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 zeroSo , 0 =kQ/3 +3k×10-9/11Q/3 = -3×10-9/11So Q= -3/11 × 10-9 C

7 months ago
Samyak Jain
333 Points
							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$\dpi{80} \sqrt{3}$) ] iwhich 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$\dpi{80} \sqrt{3}$) = 0Solving, we get Q =  – (9$\dpi{80} \sqrt{3}$ / 11$\dpi{80} \sqrt{11}$) x 10 – 9 C.

7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Electrostatics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions