Answer to Question #245825 in Electrical Engineering for jinbitchon

Question

Two slabs of material of dielectric strengths 4 and 6 and of

thickness 2mm and 5mm respectively are the inserted between the

plates of a parallel plate capacitor. Find how much the distance

between the plates should be changes so as to restore the potential

of the capacitor to its original value.

Answer 1

A capacitor is a gadget used to store electric charge. Capacitors have applications going from sifting static through of radio gathering to energy stockpiling in heart defibrillators. Regularly, business capacitors have two leading parts near each other, yet not contacting, like those in Figure 1. (More often than not a protector is utilized between the two plates to give partition—see the conversation on dielectrics beneath.) When battery terminals are associated with an at first uncharged capacitor, equivalent measures of positive and negative charge, +Q and – Q, are isolated into its two plates. The capacitor stays unbiased generally, however we allude to it as putting away a charge Q in the present situation. 

A framework made out of two indistinguishable, equal leading plates isolated by a distance, as in Figure 2, is known as an equal plate capacitor. It is not difficult to see the connection between the voltage and the put away charge for an equal plate capacitor, as displayed in Figure 2. Every electric field line begins a singular positive charge and finishes on a negative one, so that there will be more field lines in case there is more charge. (Drawing a solitary field line for each charge is an accommodation, in particular. We can draw many field lines for each charge, however the absolute number is relative to the quantity of charges.) The electric field strength is, subsequently, straightforwardly corresponding to Q. 

The field is relative to the charge: 

E∝Q,where the image ∝ signifies "relative to." From the conversation in Electric Potential in a Uniform Electric Field, we realize that the voltage across equal plates is 

V = Ed. 

In this way, V∝E. It follows, then, at that point, that V∝Q, and then again, 

Q∝V. 

This is valid overall: The more noteworthy the voltage applied to any capacitor, the more prominent the charge put away in it. 

Various capacitors will store various measures of charge for a similar applied voltage, contingent upon their actual qualities. We characterize their capacitance C to be with the end goal that the charge Q put away in a capacitor is corresponding to C. The charge put away in a capacitor is given by 

Q = CV. 

This condition communicates the two main considerations influencing the measure of charge put away. Those variables are the actual qualities of the capacitor, C, and the voltage, V. Revising the condition, we see that capacitance C is the measure of charge put away per volt, or 

C 

= 

Q 

V