Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

This is electric charges and fields chapter. Answer is 3rd option Explain with diagram sir.

This is electric charges and fields chapter.
Answer is 3rd option
Explain with diagram sir.

Question Image
Grade:

1 Answers

Venkat
273 Points
2 years ago
Charge at B due charge at A
F_{1}=\frac{1}{4\pi \varepsilon _{0}}\frac{(2\mu*C)(2\mu*C) }{AB^{2}} \\F_{1}=\frac{1}{4\pi \varepsilon _{0}}\frac{(4*10^{-12}) }{0.03^{2}}
Note that the force F1 is a vector
 
Charge at B due to C.
F_{2}=\frac{1}{4\pi \varepsilon _{0}}\frac{(2\mu*C)(2\mu*C) }{BC^{2}} \\F_{2}=\frac{1}{4\pi \varepsilon _{0}}\frac{(4*10^{-12}) }{0.04^{2}}
Note that the force F2 is a vector.
 
Let us take the resultant of the above two forces make an angle \theta with horizontal line BC. So,
tan\theta =\frac{F_{1}}{F_{2}} \\ \\tan\theta =\frac{0.04^{2}}{0.03^{2}} \\ \\\theta =tan^{-1}\frac{16}{9}
Hence, correct answer is option C.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free

10th iCAT Scholarship Test Registration Form