Home
Class 12
PHYSICS
How many electrons must be added to one ...

How many electrons must be added to one plate and removed from the other so as to store 25.0 J of energy in a 5.0 nF parallel plate capacitor?

Text Solution

AI Generated Solution

To find out how many electrons must be added to one plate and removed from the other to store 25.0 J of energy in a 5.0 nF parallel plate capacitor, we can follow these steps: ### Step 1: Use the formula for energy stored in a capacitor The energy (U) stored in a capacitor is given by the formula: \[ U = \frac{Q^2}{2C} \] where \(U\) is the energy, \(Q\) is the charge, and \(C\) is the capacitance. ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SAMPLE QUESTION PAPER 2019

    CBSE COMPLEMENTARY MATERIAL|Exercise Section B|8 Videos

  • ELECTROSTATICS AND CURRENT ELECTRICITY

    CBSE COMPLEMENTARY MATERIAL|Exercise NUMERICALS|49 Videos

  • UNIT–III & UNIT–IV MAGNETIC EFFECTS OF CURRENT AND MAGNETISM & E.M.I. AND ALTERNATING CURRENT

    CBSE COMPLEMENTARY MATERIAL|Exercise (NUMERICALS)|57 Videos

Similar Questions

Explore conceptually related problems

a. How many excess electrons must be added to one plate and removed from the other to give a 5.000 nF parallel plate capacitor 25.0 J of stored energy ? b. How could you modify the geometry of this capacitor so that can store 50.0 J of energy without changing the charge on its plates?

Derive an expression for the energy stored in a charged parallel plate capacitor.

Knowledge Check

  • Consider a parallel plate capacitor originally with a charge q_0 capacitance C_0 and potential difference Delta V_0 There is an electrostatic force of magnitude F_0 between the plates and the capacitor has a stored energy U_0 . The terminals of the capacitor are connected to another capacitor of same capacitance and charge. Later the dielectric slab is removed. WHile the slab is being removed.

    A
    Charge on second capacitor increases
    B
    Charge on second capacitor decreases
    C
    Charge on second capacitor is constant
    D
    Can't be said

  • A plate carries a charge of -3.0 muC , while a rod carries a charge of +2.0 mu C . How many electrons must be transferred from the plate to the rod, so that both objects have the same charge ?

    A
    `6.3 xx 10^(12)` electrons
    B
    `1.2 xx 10^(13)` electrons
    C
    `8.0 xx 10^(12)` electrons
    D
    `1.6 xx 10^(13)` electrons

    • Similar Questions

    Explore conceptually related problems

    In what form of field, is the energy of a capacitor stored? Derive an expression for energy of a parallel plate capacitor.

    Two identical copper spheres are separated by 1m in vacuum.How many electrons would have to be removed from one sphere and added to the other so that they now attract each other with a force of 0.9N ?

    How would you modify this capacitor so that it can store 50.0 J of energy without changing the charge on. its plates?

    Compare the expression for magnetic energy density with electrostatic energy density stored in the space between the plates of a parallel plate capacitor.

    Separation between the plates of a capacitor is increased from d to 2d while the battery remains connected with the plates. How will the stored energy in the capacitor change?

    What happens to the stored electric potential energy and the electric field between the plates of a charged isolated parallel plate capacitor when the plate separation is increased?

    Home
    Profile