Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C

Question

Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

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Khánh Gia 5 months 2021-08-27T12:25:07+00:00 1 Answers 0 views 0

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    2021-08-27T12:26:22+00:00

    Answer:

    The voltage is  V  =   418.60 \ Volts  

    Explanation:

    From the question we are told that

        The area of the both plate is  A =  7.00 *10^{-3} \ m^2

        The distance between the plate is d = 4.80*10^{-4}\ m

         The magnitude of the charge is  q = 5.40 *10^{-8} \ C

       

    The capacitance of the capacitor that consist of the two plates is mathematically represented as

            C =  \frac{\epsilon _o A}{d}

    Where \epsilon_o is the permitivity of free space with a value  e = 8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

    So

           C =  \frac{8.85*10^{-12} *  (7* 10^{-3})}{ 4.8*10^{-4}}

            C = 1.29 *10^{-10} \ F

    The potential difference between the plate is mathematically represented as

          V  =  \frac{ Q}{C }

         V  =  \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}

         V  =   418.60 \ Volts

       

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