×

#### 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
```
A circular ring carries a charge q the variation if electric field is x measured from centre along axis the maximum electric field on the axis is

```
5 months ago

## Answers : (1)

Arun
24738 Points
```							With C = Q/(4πε₀) the field along the axis is: E = Cx/(x² + R²)^(3/2) (equation 1) This equation can be derived with calculus - see derivation in link for example (short pdf document) When x E ≈ Cx/(R²)^(3/2) . .≈ (C/R³)x (C/R³) is a constant, so this means E varies as (is proportional to) x approximately: E∝x  _________________ The maximum electric field occurs at x = x₀ = R/√2 (see the link). Subsititute x = R/√2 in equation 1 giving: Emax = C(R/√2)/((R/√2)² + R²)^(3/2) If you do a bit of algebra to simplify this, it should give the required answer.
```
5 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Physical Chemistry

View all Questions »

### Course Features

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

### Course Features

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