 ×     #### 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
```
two identical balls each having a charge of 2×10-7 and a mass of 100g are suspended from a common point by two insulating strings each 50 cms long. The balls are held at a separation 5 cm apart and then released . find the 1]the electric force on one of the charged balls.2]the components of the resultant force on it along and perpendicular to the string.3]the tension in the string4]the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release

```
11 years ago

```							Hi
Find the angle  formed by the strings by using sine law (trigonometry) : a /sin a = b/sinb = c/sinc . a=5cm , b= c=50 cm , angle b = angle c
1. u can find the force on the ball by simply using the coulombs law f =kq1q2/r2 where r = 5cm .
2. now since u know the angles, find the resultants in the respective directions by taking components of the force in that direction
3. Draw force body diagram of the ball and balance the forces along the string because it is unstretchable, hence u can find the tension in the string at that instant
4. after that , calaculate acceleration perpendicular to the string by taking into account the components of tension , wt and eletrostatic force in that direction,.
```
11 years ago
```							The balls and and the suspension point form an isosceles triangle with the side length s = 50cm = 0.5m and the base line b = 5cm = 0.05m The angle β between the strings at the suspension point can be evaluated by basic trigonometry. The angle bisector of β cuts the isosceles triangle into two right triangles with hypotenuse s and b/2 as cathetus adjacent to β/2. Hence sin(β/2) = b/2 / s = 0.05 → β/2 = 2.87° (you need this angle later) (a) The electrostatic force on the ball is given by Coulomb's law: F_e = k · q₁ · q₂ / r² = 8.998·10^9 Nm²/C²· · (2·10^(-7)C)² / (0.05m)² = 0.144N This force is a repulsive force acting along the base of the triangle b. (b) The second force acting on the balls is gravity: F_g = m · g = 0.1kg · 9,81= 0.981N It acts downwards and perpendicular to the electrical force. To split the two forces into components with respect to the direction of the string, drop a perpendicular to the direct line of the string. The two components and the force vector will form a right triangle with the components as legs. These are given by alongstring: F_s = F·cosα perpendicular to string: F_p = F·sinα where α is the angle between the force and the string. From the sketch you will see: α_g = β/2 α_e = 90° - β/2 You get the components of the resultant force by adding the components of the gravity force and electric force: F_rs = F_g·cos(β/2) + F_e·cos(90°-β/2) = F_g·cos(β/2) + F_e·sin(β/2) = 0.987N F_rp = F_g·sin(β/2) + F_e·sin(90°-β/2) = F_g·sin(β/2) + F_e·cos(β/2) = 0.987N = 0.193N (c) Sketch sketch a ball with the three forces (as vectors) resulting force Fr, tension T and inert force m·a (which is a fictitious force). The balance of these forces leads to the equation of motion: F_r + T - m·a = 0 → m·a = F_r + T On the one hand, the tension is a force which can only along the direction of the string. On the other hand, the acceleration vector can show in the direction of motion. Since the ball can only move like a pendulum, this direction is perpendicular to the string. So if you break the vector equation above into components with respect to the string direction you will get: T = -F_rs m·a = F_rp The tension in the string: T = -F_rs = -0.987N The negative sign indictaes the force acting in diffrent direction. (d) m·a = F_rp → a = F_rp / m = 0.193N / 0.1kg = 1.93m/s²
```
2 years ago 605 Points
```							Dear student,Please find the solution to your problem. The balls and and the suspension point form an isosceles triangle with the side length s = 50cm = 0.5m and the base line b = 5cm = 0.05m The angle β between the strings at the suspension point can be evaluated by basic trigonometry. The angle bisector of β cuts the isosceles triangle into two right triangles with hypotenuse s and b/2 as cathetus adjacent to β/2. Hence sin(β/2) = b/2 / s = 0.05 → β/2 = 2.87° (you need this angle later) (a) The electrostatic force on the ball is given by Coulomb's law: F_e = k · q₁ · q₂ / r² = 8.998·10^9 Nm²/C²· · (2·10^(-7)C)² / (0.05m)² = 0.144N This force is a repulsive force acting along the base of the triangle b. (b) The second force acting on the balls is gravity: F_g = m · g = 0.1kg · 9,81= 0.981N It acts downwards and perpendicular to the electrical force. To split the two forces into components with respect to the direction of the string, drop a perpendicular to the direct line of the string. The two components and the force vector will form a right triangle with the components as legs. These are given by alongstring: F_s = F·cosα perpendicular to string: F_p = F·sinα where α is the angle between the force and the string. From the sketch you will see: α_g = β/2 α_e = 90° - β/2 You get the components of the resultant force by adding the components of the gravity force and electric force: F_rs = F_g·cos(β/2) + F_e·cos(90°-β/2) = F_g·cos(β/2) + F_e·sin(β/2) = 0.987N F_rp = F_g·sin(β/2) + F_e·sin(90°-β/2) = F_g·sin(β/2) + F_e·cos(β/2) = 0.987N = 0.193N (c) Sketch sketch a ball with the three forces (as vectors) resulting force Fr, tension T and inert force m·a (which is a fictitious force). The balance of these forces leads to the equation of motion: F_r + T - m·a = 0 → m·a = F_r + T On the one hand, the tension is a force which can only along the direction of the string. On the other hand, the acceleration vector can show in the direction of motion. Since the ball can only move like a pendulum, this direction is perpendicular to the string. So if you break the vector equation above into components with respect to the string direction you will get: T = -F_rs m·a = F_rp The tension in the string: T = -F_rs = -0.987N The negative sign indictaes the force acting in diffrent direction. (d) m·a = F_rp → a = F_rp / m = 0.193N / 0.1kg = 1.93m/s² Thanks and regards,Kushagra
```
3 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