Determining charge from electric potential

In summary, the problem involves a metal sphere with radius ra = 1.20 cm and a hollow, metal, spherical shell with radius rb = 9.00 cm. Charge +q is placed on the inner sphere and charge -q on the outer spherical shell, with the potential difference between the spheres set to 500 V. The equation used to calculate q is ΔV=q/K [(1/ra)-(1/rb)], where K=1/4∏ε0. After solving for q, the correct value is found to be q= 6.23*10^10 C. The mistake in the initial solution was forgetting to include the constant K in the equation.
  • #1
rmcgovern
8
0

Homework Statement


A metal sphere with radius ra = 1.20 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.00 cm. Charge +q is put on the inner sphere and charge -q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 500 V, with the inner sphere at higher potential. Calculate q.


Homework Equations



ΔV=q/K [(1/ra)-(1/rb)]


The Attempt at a Solution



Solving for q from the eqn above I have reached q= ΔV/[(1/ra)-(1/rb)] * K. By simply plugging in the numbers given I reached a soln of q= 6.23*10^10 C. This answer seemed far too large to be right, and indeed it is not correct. Is this the proper equation to use in this situation? I double checked the math and it appears everything is right, does anyone see where I may have gone wrong? Any help is greatly appreciated.
 
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  • #2
rmcgovern said:
ΔV=q/K [(1/ra)-(1/rb)]

Hi rmcgovern. Welcome to PF!

Did you put the constant K in the correct location in your equation for ΔV?
 
  • #3
ah! I forgot that K=1/4∏ε0. a rather trivial error. thank you for your help.
 

1. How is electric potential measured?

Electric potential is measured using a device called a voltmeter. It measures the difference in electric potential between two points, also known as the voltage.

2. What is the relationship between electric potential and charge?

Electric potential is directly proportional to the charge of an object. This means that as the charge increases, the electric potential also increases.

3. Can electric potential be negative?

Yes, electric potential can be negative. This occurs when the electric potential at a point is lower than the reference point, usually taken to be at infinity. Negative electric potential indicates that work must be done to move a positively charged particle from infinity to that point.

4. How is charge calculated from electric potential?

Charge can be calculated from electric potential using the formula Q = CV, where Q is the charge, C is the capacitance of the object, and V is the electric potential. The capacitance is a constant that depends on the geometry and material of the object.

5. What is the unit for electric potential?

The unit for electric potential is the volt, which is represented by the symbol V. It is equivalent to one joule per coulomb (J/C), where joule is the unit for energy and coulomb is the unit for charge.

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