# Example: Two point charges are separated by a distance of 20 cm. The two charges q 1 and q 2 have magnitudes of 5  C and 12  C, respectively. a)What.

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Example: Two point charges are separated by a distance of 20 cm. The two charges q 1 and q 2 have magnitudes of 5  C and 12  C, respectively. a)What is the electric potential at a point 30 cm from each charge? b)How much energy is required to bring in a third charge q 3 with a magnitude of 10  C to a distance of 30 cm from each charge? a) q1q1 q2q2 20 cm q3q3 30 cm b) P

Part b) of the previous problem could also be solved using electric fields. Let us look at how this might be done.  1 =  2 Isosceles triangle q1q1 q2q2

Two test charges are brought separately into the vicinity of a charge + Q. First, test charge + q is brought to point A a distance r from + Q. Next, + q is removed and a test charge +2 q is brought to point B a distance 2 r from + Q. Compared with the electrostatic potential of the charge at A, that of the charge at B is 1. greater. 2. smaller. 3. the same.

Two test charges are brought separately into the vicinity of a charge + Q. First, test charge + q is brought to a point a distance r from + Q. Then this charge is removed and test charge – q is brought to the same point. The electrostatic potential energy of which test charge is greater: 1. + q 2. – q 3. It is the same for both. Gain of energy Loss of energy

We have found the electric potential at a point from our knowledge of the electric field at that point. Can we determine the electric field from the electric potential? Yes we can. Let us begin with our expression relating the electric potential to the electric field. If we assume that the electric field is only in the x-direction, we can simplify the previous expression. Similarly for the other directions The electric field is proportional to the rate of change of the electric potential, with respect to position. This would be the slope of a plot of the electric potential with respect to position. For radially symmetric systems you may use a similar expression. For polar, cylindrical or spherical symmetry.

To see how this works let us look at out expression for the electric potential of a point charge. Electric field of a point charge! If we have an electric potential in a radial direction and want to find it in each of the coordinate directions we can use partial derivatives. Partial derivative are nearly identical with full derivatives, with the exception that all variables besides the one we are interested in are considered constant. Symbol for partial derivative Also remember: Example:

Electric potential due to a continuous charge distribution We have an expression for the potential due to a point charge. We may also find it necessary to modify this expression to be able to find the electric potential due to a continuous charge distribution. The modification is similar to what was done with the electric field. For small piece of the total potential, due to a small part of the total charge For the total potential due to the continuous charged object  V is not the potential difference in this case it is a small amount of the total potential

Example: Find the potential at point P, a distance x from the center of the ring, due to the charged ring of radius a and total charge Q. r x x r in terms of known quantities a and x We can work with dq instead of looking at charge density mainly because electric potential is a scalar quantity.

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