Home
Class 12
PHYSICS
A bodyX has+2 coulomb of charge and body...

A bodyX has`+2` coulomb of charge and bodyY has `-2` coulomb of charge . Can we say that X has more charge as compared to Y ?

Text Solution

AI Generated Solution

To determine whether body X has more charge compared to body Y, we need to analyze the charges of both bodies: 1. **Identify the Charges**: - Body X has a charge of +2 coulombs. - Body Y has a charge of -2 coulombs. 2. **Understanding Charge Convention**: - The convention of electric charge states that positive and negative charges are of opposite nature. This convention was established by Benjamin Franklin, who assigned positive to the charge on a glass rod rubbed with silk and negative to the charge on a plastic rod rubbed with wool. ...
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE ENGLISH|Exercise TRY YOURSELF|33 Videos
  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT(SECTION-A) Objective Type Question|45 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D)|10 Videos
  • ELECTROMAGNETIC INDUCTION

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT(SECTION -D) Assertion-Reason type Question)|15 Videos

Similar Questions

Explore conceptually related problems

A positively charged body has :

A body has -80 micro coulomb of charge. Number of additional electrons in it will be

Calculate the number of electrons consituting one coulomb of charge.

How many electrons are there in one coulomb of negative charge?

An object is charged when it has a charge imbalance, which means the

A body has a positive charge of 8xx10^(-19)C . It has :

How much energy is given to each coulomb of charge passing through a 6 V battery ?

How many electrons are there in on coulomb of negative charge?

The charge of an electron is 1.6 xx 10^(-19) coulombs. What will be the value of charge of Na^(+) ion

Can 8.6 xx 10^(-19) Coulomb of charge be given to a conductor? Explain.