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A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now?

In case the negative plate was also given a charge -Q then the answer would be simply V+Q/C but in this case I'm confused.Can someone please guide me or give some hints how to find potential difference in case only one plate is charged?

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  • \$\begingroup\$ This looks to be a homework question. Please explain your work so far and what concepts you are applying. Then point out where you are lacking in understanding. \$\endgroup\$ – user2943160 Aug 30 '16 at 2:41
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    \$\begingroup\$ This is probably more a Physics question. In normal circuit operation the two plates of a capacitor are always charged with equal charges of opposite sign. \$\endgroup\$ – Lorenzo Donati Aug 30 '16 at 2:50
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    \$\begingroup\$ Actually, upon returning to my browser and re-reading, this is an extremely poorly worded question. \$\endgroup\$ – Daniel Aug 30 '16 at 3:08
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    \$\begingroup\$ Hint: Nowhere in the question you quoted does it say the other plate is not charged. Think about Kirchoff's current law and what it means about what happens to the second plate when the first plate is charged. \$\endgroup\$ – The Photon Aug 30 '16 at 3:48
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    \$\begingroup\$ @ThePhoton I think you are mistaken.On deposition Q charge on one plate the charge gets distributed between the two surfaces in such a way to make Gauss Law valid.The other plate may not not be given any extra charge at all.Anyway I found out the answer from several books that the P.D will be V+Q/2. \$\endgroup\$ – user72436 Aug 31 '16 at 2:59
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Take it as a physics problem: charge the capacitor as a whole, then also place the usual charge-difference across the two plates, such that one plate now has a total charge of zero.

Heh. Probably NOT what the textbook question intended! But not impossible.

First, suppose we have an isolated metal sphere with 1pF capacitance (the 1pF measured to a plate at infinity,) then we place 1uC of positive charge upon the sphere. By Q=CV the surface voltage will be 1 megavolt (voltage measured to a plate at infinite distance.) A one-plate capacitor? Close, but not exactly, since charge is conserved, so there's always an opposite plate at infinite distance.

Next, we bring an identical but uncharged metal sphere very close to the first one, so the capacitance between the two spheres is fairly large, say 10,000pF or 0.01uF. That uncharged sphere will now display an induced charge-pair, with the negative half attracted to the side facing the original pos-charged sphere, and the positive half repelled to the side facing away from the pos-charged sphere. (The extra sphere still has zero average charge, but it's now composed of regions, having equal and opposite charge.)

This 2-sphere capacitor ends up having 0.5 microcoulomb on either side of the narrow gap (pos on one plate, neg on the other.) By Q=CV, the voltage on this 0.01uF capacitor is 50V, measured between the two spheres. But 1MV on the whole capacitor, either sphere measured to distant ground.

The trick here is that, with most real-world capacitors, the capacitance between the plates is extremely large when compared to the few-picofarads capacitance measured from each plate to distant ground. In the above example it's 10,000 to 1 ratio. Because of the large difference, this causes the charge-separation found on an uncharged capacitor plate to conveniently be very close to half the charge of the charged plate. Then, if we try to charge up just one of the plates, the voltage between the two plates will be almost insignificant when compared to the voltage between either plate and ground.

Confusing. Needs animated diagrams!

Here's another trick to clarify things: don't use two spheres. Instead, visualize two solid metal hemispheres, flat faces almost together with a thin dielectric layer clamped between their faces. It looks like a solid sphere, but with a thin slot sliced between. I call it the "Engineer's capacitor." The two separate spheres is a "physicist's capacitor."

Next, put a 1 micro-coulomb positive charge on just one hemisphere, leaving zero charge on the other. The positive hemisphere will end up with 1/2 uC of positive charge on the flat face, and 1/2 uC of positive charge on it's curved hemisphere surface, for a total charge of 1 uC. The other uncharged hemisphere will have 1/2 uC of negative charge on the flat face, and 1/2uC of positive charge on the curved surface, for a total net charge of zero. In other words, the two flat faces have pos and neg 1/2uC, while each of the curved hemispheres gets pos 1/2 uC. All together, the device behaves like a single metal ball with 1uC total charge distributed almost uniformly across its outer curved surface ...but also with a 10,000pF capacitance created by the narrow slot. (If the complete sphere had 1pF capacitance to ground, then yes, the 50V between the hemispheres does slightly alter the megavolt of surface potential on each individual hemisphere.)

Conclusion: it is possible to have a capacitor with one charged plate and one neutral plate. We'd need to know the value of the capacitor, C1, and also the picofarad value measured between the capacitor and ground, C2. Then we place a tiny voltage between the capacitor plates, while also using an immense voltage to charge up the capacitor as a whole. The ratio of the two voltages must be 2*C1/C2, or 20,000/1 in the above hemispheres example.

Finally, if you wanted to actually perform a desktop experiment using small metal spheres, you'd be wise to use nanocoulombs and kilovolts rather than microcoulombs and megavolts. Rule of thumb: a microcoulomb is the amount of electric charge on human body connected to 10KV wrt ground, since humans have roughly 100pF wrt ground. Human-sized kilovolt capacitor plates are too unweildy :)

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Let's say,if you apply a +Q to the capacitor,Then equal and opposite potential will be formed the negative plate of the capacitor(opposite plate of capacitor).If the charge in the capacitor is increased,then automatically,the potential (V) of the capacitor will also increase (Q is directly proportional to V),given that the other terminal of the capacitor should be connected to ground or specific potential to get the potential difference across the two plates of capacitor .

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