Question

A 0.16 pF parallel - plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by

Answer 1

The value is
`V_d  = 5 \ V`

Step-by-step explanation:

From the question we are told that

The capacitance of the capacitor is
`C =  0.16 \  pF  =  0.16*10^(-12) \ F`

The voltage is
`V  =  10\ V`

The number of electrons present on the negative plate is
`N_e  =  1.00 *10^(7) \  electrons`

Generally the charge on the positive plate at 10 Volt is mathematically represented as

`Q_a =  C *  V`

=>
`Q_a  =  0.16*10^(-12) *  10`

=>
`Q_a  =  0.16*10^(-11) \  C`

Now the charge on the plate when the electron where placed is evaluated as

`Q_b  =  Q_a  + ( N_e * e)`

Where e is the charge on each electron with a value
`e =  1.60 *10^(-19) \ C`

=>
`Q_b  =  0.16*10^(-11) + (1.0*10^7 *  1.60*10^(-19))`

=>
`Q_b  = 3.2 *10^(-12 ) \  C`

Generally the voltage between the two plate is evaluated as

`V_d  =  ( Q_b  -  Q_a  )/( 2 * C )`

=>
`V_d  = ( 3.2*10^(-12) -  0.16*10^(-11))/(2 *  0.16 *10^(-12))`

=>
`V_d  = 5 \ V`