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

The  value is  

Explanation:

From the question we are told that

    The  capacitance of the capacitor is

     The  voltage is  

    The number of electrons present on the negative plate is  

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

     =>  

     =>    

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

Where e  is the charge on each electron with a value  

     =>    

      =>  

Generally the voltage between the two plate is evaluated as

=>    

=>      

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explanation: