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An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

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okpalawalter8

Answer:

[tex]p.d' = 37.6 V[/tex]

Explanation:

From the question we are told that:

Potential difference [tex]p.d=18.8V[/tex]

New Capacitor [tex]C_1=C_2/2[/tex]

Generally the equation for Capacitor capacitance is mathematically given by

[tex]C=\frac{eA}{d}[/tex]

Generally the equation for New p.d' is mathematically given by

 [tex]C_2V=C_1*p.d'[/tex]

  [tex]p.d' = 2V[/tex]

 [tex]p.d'= 2 * 18.8[/tex]

 [tex]p.d' = 37.6 V[/tex]