Reference no: EM131181512

Electricity Assignment-

Text: Randall D. Knight, Physics for Scientists and Engineers: A Strategic Approach

Hand in: Submission slot outside rm. 67-341, level 3 Priestly bld.

Question 1 -

The figure shows a square plate of side-length L charged with uniform surface charge density η. It is centered at x = y = z = 0 and orientated in the z = 0 plane. Your task is to determine the z-component of the electric field at the point r(x, y, z) = r(x, 0, d), offset along the x-axis and a height d above the plane of the plate.

Our strategy is to break the plate into a set of thin strips, with the i^th strip having a thickness Δx_i and position x_i along the x-axis (see figure). The electric fields of the strips can then be individually determined, and summed up via an integral to obtain the total electric field. In the y = 0 plane bisecting the plate, the electric field magnitude of strip i is

E_i = (|η|LΔx_i/4πε₀r_i)[r_i² + (L²/4)]^-1/2,

where r_i is the distance from the centre of the strip to the point r.

Part A-

(a) Draw two sketches of the system in the y = 0 plane (side view), the first showing the electric field lines from the plate, and the second showing the total electric field vector at point r as well as its z-component.

(b) Using your field vector sketch along with the expression for E_i above show that, in the y = 0 plane, the z-component of the electric field from strip i is

(E_i)_z = (ηLdΔx_i/4πε₀r_i²)[r_i² + (L²/4)]^-1/2.

(c) Now determine the z-component of the total electric field of the square plate in the y = 0 plane. You may find that the integral given over the page is useful.

Part B-

In the limit that d << L, the result of the previous question can be expressed as:

(E_plate)_z = (η/2πε₀){arctan((L/2d)(L-2x/√(L²+(L-2x)²))) + arctan((L/2d)(L+2x/√(L²+(L+2x)²)))},

where arctan is the inverse tan function.

(d) Use this expression and symmetry arguments to show that when d << L the total electric field above the centre of the plate (i.e., when x = y = 0) equals the field of an infinite plate.

(e) Now consider the case where x ≈ L/2, i.e., where the point r is near the edge of the plate along the x-axis. Show that when both d and |L - 2x| are much less than L the electric field is reduced in magnitude compared to the centre of the plate (x = 0) by the factor R = 1 /2 - arctan[(x - (L/2))/d]/π.

(f) What does the above results tell you about the spatial extent of the fringing field (the non-ideal field that exist near the edge of the plates) of a non-infinite parallel plate capacitor?

Question 2-

Part A-

Three positive charges, each of magnitude Q, are fixed as shown in the figure. Each charge is a distance of L from the origin, with two of the charges lying on the x-axis at x = ±L, and one charge lying on the y-axis at y = -L.

(a) It is possible to determine the electric field along the y-axis using two approaches:

• A direct calculation using the electric field for a point charge, and the principle of superposition;
• By first calculating the electric potential, and from this, the electric field.

Discuss the advantages and disadvantages of these two approaches for this charge configuration.

(b) Derive an expression for the electric potential along the positive y-axis.

(c) A positive test charge, also of magnitude Q, is released from the origin. Discuss the motion of the charge after it is released.

(d) Suppose the four charges (the three fixed charges and the test charge) are all protons and L is 15 mm. Calculate the speed of the test charge when it is infinitely far from the fixed charges.

Part B (Advanced)-

Calculate the electric field for the charge distribution from Part A, both directly by superposition, and using the potential that you calculated above. Comment on what you find.