Part One Consider a uniformly charged sphere of radius R and charge Q. Let us no
ID: 3161605 • Letter: P
Question
Part One
Consider a uniformly charged sphere of radius R and charge Q.
Let us now place a test charge qo near the sphere.
a. The potential energy of the system containing the charged sphere and the test charge is equal to the work done by an external force to bring the test charge from a reference point to its final position What is an appropriate reference point to use and why?
b. the potential energy of the system were negative, would the electric field of the sphere have done positive or negative work on the test charge as it was brought in from the reference point? Would correspond to an attractive force or a repulsive force between the sphere and the test charge?
c. if the sphere was brought in from the reference point to the test charge instead of the test charge brought in from the reference point to the sphere, would the potential energy of the system change? Would the potential energy of the system be different if the sign of the test charge changed?
d. considering the answer you gave above, why is potential, rather than potential energy, a better way to describe the electrical effects caused by the sphere?
Let's remove the test charge and calculate the potential of the charged sphere.
e. Use Gauss's Law to find the electric field everywhere inside and outside the sphere
f. Integrate the electric field to find the potential outside the sphere at r=2R. at the edge of the sphere at r = R, and inside the sphere at r =R/2.
oads/APSC%20178%20-%20Tutorial%204 pdf Temp Re submit handwritten solutions to your TA during the tutorial sestion. Show your work and demonstrate the steos you used to solve each problem Please include your name, student number, and tutorial section on Your submitted work. Part One Consider a uniformly charged sphere of radius R and charge Q Let us now place a test charge near the sphere. a The potential energy of the system containing the charged sphere and the test charge is equal to the work done by an esternal force to bring the test charge from a reference point to its final position What is an appropriate reference point to use and whv? b. the potential energy of the system were negative, would the electric field of the sphere have done positive or negative work on the testcharge as a was brought in forn the reference point would correspond to an attractive force or a eepulsive force between the sphere and the test charge? c if the sphere was brought in from the reference point to the test charge instead of the test charge brought in from the reference point to the uphere, would the potential energy of the system charee? Would the potential energy of the system be dinerent fthe sign of the test charge changed? d considering the answer vou gave above why is potential, rather than potential energy, a better way to describe the electrical effects caused by the sphere? Let's remove the test charge and calculate the potential of the charged sphere. e. Use Gauss's Law to find the electric field everywhere inside and outside the sphere f Integrate the electric field to find the potential outside the sohere at r 2R. at the edge of the sphere at r R, and inside the sphere at r RR.Explanation / Answer
a.) Reference point would be infinity because at infinity potential energy is zero and as we moving it towards charge body potential energy gets increase and that increase in potential energy is work done .
b.)Electric field has done negative work done if the potential energy of the system is negative
work done = change in potential energy
it would correspond to be attractive force as potential energy is negative
U = KQq/r
c.)Potential energy will remain Same
U = KQq/r
yes the Sign of the potential energy would be differen if the sign of test charge changes.
d.) The change in voltage is defined as the work done per unit charge against the electric field. In the case of constant electric fieldwhen the movement is directly against the field, this can be written
Vf - Vi = - Ed
e.)r>R
Electric field outside the sphere = KQ/r^2
r <R
Electric field inside the sphere = 0
f)
at r=2R
E = KQ/4R^2
at r = R'/2
E = 0
The electric field is seen to be identical to that of a point charge Q at the center of the sphere. Since all the charge will reside on the conducting surface, a Gaussian surface at r< R will enclose no charge, and by its symmetry can be seen to be zero at all points inside the spherical conductor