1. A charged insulating sphere radius of b has a non-constant volume charge dens

Question

1. A charged insulating sphere radius of b has a non-constant volume charge density of = 11r^(2/3). Find the electric field magnitude in each region in terms of (outside the object and inside the inside the object). Sketch a graph of the electric field magnitude as a function of radius for r = 0 to r = 2b. Make sure it is roughly to scale and label the magnitude of the field at r = b and 2b. Then find the electric potential in each region in terms of , and sketch a graph of the electric potential as a function of radius for r = 0 to r = 2b. Make sure it is roughly to scale and label the potential at r = 0, b, and 2b. Your reference point for the potential is that V = 0 at r = .

2. A circular arc of charge having radius a is centered on the +x-axis and subtends an angle of theta both above and below the x-axis. If the arc has a linear density of lambda = lambda(subscript)0 cos theta, what is the electric field at the center of the arc? Give both magnitude and direciton (assume lambda(subscript)0 is positive). You may need the identity cos^2 theta = (1 + cos (2 times theta) / 2).

3. In a certain region of space, the electric potential is defined by the function:

V (x, y, z) = [12(y^3)(z^4) / (x^(5/2))] - [(1/8)(x^2)(y^-2)(z^3)] + [4(x)(z^2)]

a) Find the general expression for the electric field in component form.

b) What is the magnitude of the electric field at the point (4, 1, 2)?

4. A 20.0 ohm resistor has a rectangular cross-section with length, I, that is constant and a width, w, that varies linearly from one end to the other of the resistor (that is dw/dx is constant). At one end, the rectangle is 3.40 cm wide, and at the other end, it is 8.55 cm wide. The long side of the rectangle has a width of 9.10 cm. The resistor is 8.45 cm deep. What is the resistivity of the material composing it? (Note: On the figure, side lengths are indicated, and the figure is not to scale. Assume that the shape has no effect on the resistance.)

1. A charged insulating sphere radius of b has a non-consto the electric field magnitude in e object). Sketch a graph of the electric field itude of the field at dle sure it is roughly to scale and label the a otential in each region i has a non-constant volume charge density of p-11a?'s. Find for r-0 to r =20 Make nitude of the field atrsb and 2b. Then, find the electric ,b, and 2b. Your- a (outside the object and inside the inside the ction of radius for r = aph of the electric field magnitude as a function of ra tric field nm f the electric potential as a function of radi us region in terms of a, and sketch a graph of the electric potential as a funct for r = 0 t reference point for t or20-Make sure it is roughly to scale and label the potential at r=0 he potential is that V = 0 at r de an angle of both

Explanation / Answer

given radius, = b
charge density, rho = 11*alpha*r^2/3

a. Electric field inside sphere at radius
from gauss' law, E*4*pi*r^2 = qin/epsilon
now to find q in

consider a thin shell in the sphere at radius r, thickness dr
then charge on this shell = 4*pi*r^2*11*alpha*r^2/3 *dr = 44*pi*alpha*r^8/3 *dr
so, charge upto radius r = 44*pi*alpha*r^11/3 [ obtained by integrating the previous expression]

so, inside sphere
E = 44*pi*alpha*r^11/3 / epsilon*4*pi*r^2 = 11*alpha*r^5/3 / epsilon

outside sphere
E*4*pi*r^2 = qnet/epsilon
qnet = 44*pi*alpha*b^11/3
E = 11*alpha*b^11/3 / epsilon*r^2

E(r = b) = 11*alpha*b^5/3 / epsilon
E(r = 2b) = 11*alpha*b^5/3 / epsilon*4

b. now, E = -dV/dr
so, V = -integrate (Edr)
inside sphere
E = 11*alpha*r^5/3 / epsilon
V = 11*alpha*r^8/3 / epsilon

outide sphere
E = 11*alpha*b^11/3 / epsilon*r^2
V = 11*alpha*b^14/3 / epsilon*r^2

V(b) = 11*alpha*b^8/3 / epsilon
V(2b) = 11*alpha*b^8/3 / epsilon*4

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