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  • Stuck here, help me understand: The distance between the plates of a parallel plate condenser is 4mmand potential difference is60 volts. If the distance between the plates is increased to 12 mm, then

The distance between the plates of a parallel plate condenser is 4mm and potential difference is60 volts  . If the distance between the plates is increased to 12 mm , then

  • Option 1)

             The potential difference of the condenser will become 180 volts 

  • Option 2)

             The P.D. will become 20 volts

  • Option 3)

    The P.D. will remain unchanged

  • Option 4)

    The charge on condenser will reduce to one third

 

Answers (1)

best_answer

As we have leraned

Parallel Plate Capacitor -

C=\frac{\epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 For capacitor \frac{V_1}{V_2}=\frac{d_1}{d_2} ÞV_2= \frac{V_1d_2}{d_1}= \frac{60*12}{4}= 180 v

 


Option 1)

         The potential difference of the condenser will become 180 volts 

Option 2)

         The P.D. will become 20 volts

Option 3)

The P.D. will remain unchanged

Option 4)

The charge on condenser will reduce to one third

Posted by

Avinash

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