
Work done in moving $5\,C$ charge across the ends of conductor is $100\,J$ If the potential at the end of conductor is $10\,V$ find the potential at the other end of this conductor.
Answer
493.2k+ views
Hint: In order to solve this question we need to understand work done and potential differences. So Potential difference is the amount of work done in moving a test charge from infinity to a point against the electric field of charge. Also potential at infinity conventionally taken to be zero. Work done is the amount of potential energy stored in charge when it comes from infinity to a certain point against the electric field of another charge.
Complete step by step answer:
According to the question, Quantity of charge moved is $q = 5C$.
Work done in moving charge be $W = 100J$.
Let the two ends of the conductor be marked as A and B.
So Potential at A is ${V_A} = 10V$
And let the potential at end B is ${V_B}$.So using potential difference relation with work done we get,
$\Delta V = \dfrac{W}{q}$
Putting values we get,
$\Delta V = \dfrac{{100}}{5}V$
$\Rightarrow \Delta V = 20V$
Using $\Delta V = {V_B} - {V_A}$
We get, ${V_B} - {V_A} = 20V$
Putting values we get ${V_B} = (20 + 10)\,V$
$\therefore {V_B} = 30\,V$
So potential at the other end is ${V_B} = 30\,V$.
Note: It should be remembered that electric field always in direction of decreasing potential so when a charge moves from infinity to certain point then it actually moves against the electric field, and if charge is of same nature then it does some positive work against it and thus potential energy is stored in configuration. So suppose one charge in configuration is released then the two charges move exactly opposite to each other to conserve momentum and with exact opposite speed.
Complete step by step answer:
According to the question, Quantity of charge moved is $q = 5C$.
Work done in moving charge be $W = 100J$.
Let the two ends of the conductor be marked as A and B.
So Potential at A is ${V_A} = 10V$
And let the potential at end B is ${V_B}$.So using potential difference relation with work done we get,
$\Delta V = \dfrac{W}{q}$
Putting values we get,
$\Delta V = \dfrac{{100}}{5}V$
$\Rightarrow \Delta V = 20V$
Using $\Delta V = {V_B} - {V_A}$
We get, ${V_B} - {V_A} = 20V$
Putting values we get ${V_B} = (20 + 10)\,V$
$\therefore {V_B} = 30\,V$
So potential at the other end is ${V_B} = 30\,V$.
Note: It should be remembered that electric field always in direction of decreasing potential so when a charge moves from infinity to certain point then it actually moves against the electric field, and if charge is of same nature then it does some positive work against it and thus potential energy is stored in configuration. So suppose one charge in configuration is released then the two charges move exactly opposite to each other to conserve momentum and with exact opposite speed.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

