wo parallel plates are charged with +Q and –Q respectively as shown in the figure, where Q = 44 nC. The area of each plate is A = 0.011 m2. The distance between them is d = 6.6 cm. The plates are in the air. a. With the information given, write the equation of the capacitance in terms of ε0, A, d. C = b. Solve for the numerical value of C in pF. C= c. With the information given, express the potential difference across the capacitor in terms of Q and C. ΔV = d. Solve for the numerical value of ΔV in V. ΔV =
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An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.
Two parallel plates are charged with +Q and –Q respectively as shown in the figure, where Q = 44 nC. The area of each plate is A = 0.011 m2. The distance between them is d = 6.6 cm. The plates are in the air.
a. With the information given, write the equation of the capacitance in terms of ε0, A, d.
C =
b. Solve for the numerical value of C in pF.
C=
c. With the information given, express the potential difference across the capacitor in terms of Q and C.
ΔV =
d. Solve for the numerical value of ΔV in V.
ΔV =
e. If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF?
C =
f. If A is then increased to 2A, d is decreased to d/4, what is the value of the capacitance C in pF?
C =
g. What is the new value of potential difference ΔV in V with the original charge Q, given the values for A and d from part (f)?
ΔV =
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