
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
494.4k+ 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.
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