Draw diagrams and explain answers where appropriate Partial credit is available.

Question

Draw diagrams and explain answers where appropriate Partial credit is available. Solutions on separate paper. k 9 times 10^N middot m^2/C^2, e = 1.6 times 10^C, mass: electrons = 9.11 times 10^Kg, protons = 1.67 times 10^-27kg, 1 ev = 1.6 times 10^-19 J Four charged particles, two with charge +q and two with charge -q, are arranged as below. The positively charged particles lie on the x axis at x = +a and x= +2a. The negatively charged particles lie on the y axis at y = +a and y = -a. a) Derive a formula for the electric field vector, E, for points along the +x axis for x > 2a (such as at the point P shown) in terms of the variable distance x the charge q and the length a. Express the vector E in unit vector notation. Describe in words the direction of the vector E. b) Imagine that a negatively charged particle were placed at the point P and the other four charged particles remain fixed in place. In which direction would the electrostatic force that acts on the particle be directed, and what does this have to do with the equation F = qE?

Explanation / Answer

a) The electric filed strength at a point at x coordinate greater than 2a will be there due to four charges placed

Electric field due to +q present at x=2a will be (9*109*q)/(x-2a)2 and will be dirrected towards positive x axis

Electric field due to +q present at x=a will be (9*109*q)/(x-a)2 and will be dirrected towards positive x axis

since the negative charges have equal charge magnitude and placed symmetrically about x axis, their electric fiels will be such that the vertical components(components in y direction) of electric files will cancel each other. The resultant electric filed due tp the two negative charges will be directed along negative x axis and the magnitude will be equal to twice the magnitude of horizontal component of electric field due to one of the negative charges

Electric fuled due to negative charge placed at y=a is gven by

E=(9*109*q)/(x2+a2)

Component of this along x axis = [(9*109*q)/(x2+a2)]*x/sqrt(x2+a2) = (9*109*q*x)/(x2+a2)3/2

Electric field at the ppoint due to both the negatively charged particles = 2*(9*109*q*x)/(x2+a2)3/2 directed towards negative x axis

Therefore, resultant electric fiels at the point due to all the four charges

E = (9*109*q)/(x-2a)2 +(9*109*q)/(x-a)2-2*(9*109*q*x)/(x2+a2)3/2

= (9*109)*[1/(x-2a)2+1/(x-a)2)-2x/(x2+a2)3/2 ] directed towards positive x axis

b) Since the net electric field is directed towards positive x axis, the electrostatic force experienced by a negatively charged particle will be directed towards negtaive x axis. The magnitude of force on the particle will be equal to qE.

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