×

#### Thank you for registering.

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

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

The force between two charges 0.06m apart is 5N. If each charge is moved towards the other by 0.04m,then the force between them will become?


2 years ago

Susmita
425 Points
							Hey let q and Q be the charge.r=0.06 is the distance between them.So the force $f=5=\frac{qQ}{4\pi\epsilon r^2}=\frac{qQ}{4\pi\epsilon (0.06)^2}$Let us draw a straight line.Now draw 7 equidistant point on line:0,0.01,0.02,0.03,0.04,0.05,0.06.The charge q is located at 0 and Q is located at 0.06.If you move 0.04 unit from point 0 to the right then you will reach at the point 0.04.If you move 0.04 unit from the point 0.06 ,you will reach the point 0.02.So now the distance between the charges is (0.04-0.02)=0.02.So the force now is$f'=\frac{qQ}{4\pi\epsilon (0.02)^2}$So $\frac{f'}{f} =\frac{(0.06)^2}{ (0.02)^2}$Or,$f'=f\frac{(0.06)^2}{ (0.02)^2}=5 (\frac{3}{1})^2 =5*9=45N$Hope this helps

2 years 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