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
`        6Two free point charges +q and +4q are a distance apart.A third charge is placed so that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge?`
one year ago

```							The given charges q and 4q are free and they repel each other. Therefore , the third change should be such that it holds both the charges. Therefore, the third charge must be negative. Also, it should be placed in between the two charges.Let the distance between q and 4q be a. We place -q1 charge at distance x from q.For equilibrium of the system, the potential energy U of the system must be minimum.Now, U=- kqq1/x +k4q^2/a - k4qq1/(a-x)….(1)For, U to be minimum,dU/dx=0. Therefore ,dU/dx=0=kqq1/x^2-k4qq1/(a-x)^2. Then,1/x^2= 4/(a-x)^2(a-x)^2=4x^2(a-x)=+_2xa-x=2x orx=a/3……….(2).With this value of x, the potential energy of the system will be, from equation (1),-kqq1/(a/3)+4kq^2/a-4kqq1/(2/3a)=U…………..(3)Now, if we want to find potential energy, we must know the value of q1.Now, q1 can be found by equating force on q OR 4q to zero.Let us consider force on q.-9kqq1/a^2+kq(4q)/a^2=0. Then,q1=(4/9)q in value.Substituting this value of q1 in equation (3), we can find U.U will come out to be -(55/6)kq^2/a.
```
one year 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