Homework #2 Solutions Version 2 5 problems

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1 Homework # Solutions Version 5 problems Eercises 1. The figure shows an electron at the origin, and a grid marked off in nanometers. (a) what is the electric field (in component form) at the point (1 nm, nm), marked with a black star in the figure? (b) If I put a proton at that point, what is the magnitude of the force that it feels? e (1nm, nm) (a) The electric field due to a point charge is given b the formula E k q s r r r In this case, q s e C, so what we need to figure out is r. To get from the source to the target, we go in the +ˆ direction a distance of 1 nm, and in the ŷ direction a distance of nm, so r 1 nm(ˆ) + nm( ŷ). The distance between source and target is then and r r (1 nm) + ( nm) ( 5) nm m ˆr r r 10 9 m(ˆ) + ( 10 9 m)( ŷ) m 0.446ˆ 0.893ŷ (Notice that the components of unit vectors have no dimension: not meters or anthing, just pure numbers.) And so the electric field at the star is E k q s r ˆr ( N ) m C C ( (0.446ˆ 0.893ŷ) m) N(0.446ˆ 0.893ŷ) N/C( ˆ) N/C(ŷ) (b) The force on a target charge q t in an electric field is F q t E, so the force on a proton (qt C) is simpl F ( C)( N/C( ˆ) N/C(ŷ)) N( ˆ) N(ŷ) or 0.5 pn( ˆ) pn(ŷ)

2 (A piconewton is 10 1 N.) We note that this force points to the left ( ˆ) and upwards (ŷ), which is back towards the electron: eactl what we epect for oppositel charged objects. The question asks for the magnitude of the force, so we use the Pthagorean theorem: F (0.5 pn) + (41.0 pn) 45.8 pn or N.. Two positive charges, one with charge µc and one with charge q 4 µc, sit on the ais, 6 cm apart; the ais runs right between them. Find the electric field (magnitude and direction) on the ais, 4 cm to the right of the origin. 0.03m 0.03m 0.04m q The electric field due to two charges is the sum of the electric field due to each charge: E E 1 + E k r1 3 r 1 + k q r 3 r where r 1 is the vector from charge 1 to the target, and similarl from r. We re given and q. From the diagram, we see that r 1 (0.03 m)ŷ + (0.04 m)ˆ and r +(0.03 m)ŷ + (0.04 m)ˆ 0.03m 0.03m 0.04m r 1 r The length of both vectors is the same: r 1 r (0.03 m) + (0.04 m) 0.05 m. Now we solve q E k r1 3 r 1 + k q r 3 r [ 10 6 C k (5 10 m) 3 r ] 10 6 C (5 10 m) 3 r ( N m /C ) 10 6 C m 3 [ r 1 + r ] N ( ) [ (0.03 m)ŷ + (0.04 m)ˆ] + [(0.06 m)ŷ + (0.08 m)ˆ] C m N (0.03 m ŷ m ˆ) C m 17.3 MN/C ˆ MN/C ŷ N/C ˆ N/C ŷ I asked for magnitude and direction, but that usuall means I m reall looking for component form. However, the direction is largel to the right (since both charges are pushing to the right) and a little bit up (because the lower charge, being twice as big, is pushing harder). The magnitude is N/C or 18 MN/C.

3 3. In the figure, the four particles form a square of edge length a 5.00 cm and have charges nc, q 0.0 nc, q nc, and q nc. In unit-vector notation, what net electric field do the particles produce at the square s center? q a q 4 q 3 The electric field at the center of the square is the sum of the electric fields due to the four charges; and as is the case with Coulomb s Law, the trick part is to find the vector r for each. For eample, r 1 is the vector from to the center, which can be gotten b moving a distance 1 a in the ˆ direction, and then 1 a in the ŷ direction; thus r 1 1 aˆ 1 aŷ. The length of this vector is r a a 1 a, and so its unit vector is ˆr 1 r 1 r 1 1 a(ˆ ŷ) 1 (ˆ ŷ) 1 a B looking at the diagram, we see that the other three vectors all have the same length 1 a (call this R) and their vectors are r 1 a( ˆ) + 1 a( ŷ) and so ˆr 1 ( ˆ ŷ) r 3 1 a( ˆ) + 1 a(ŷ) and so ˆr 3 1 ( ˆ + ŷ) r 4 1 a(ˆ) + 1 a(ŷ) and so ˆr 4 1 (ˆ + ŷ) Thus the electric field at the center is E k r 1 k ˆr 1 + k q r (a/ 1 ) ˆr + k q 3 r 3 ˆr 3 + k q 4 r4 ˆr 4 q (ˆ ŷ) + k (a/ 1 ( ˆ ŷ) + k ) (a/ 1 ( ˆ + ŷ) + k ) (a/ 1 (ˆ + ŷ) ) k (a /) [(ˆ ŷ) + q ( ˆ ŷ) + q 3 ( ˆ + ŷ) + q 4 (ˆ + ŷ)] k a Now in this problem, [( q q 3 + q 4 ) ˆ + ( q + q 3 + q 4 )ŷ] q 3 q 4 q q 3 + q 4 [(10) ( 0) (0) + ( 10)] nc 0 q + q 3 + q 4 [ (10) ( 0) + (0) + ( 10)] nc 0 nc Thus the electric field onl has a component: the components of the fields here cancel. The field itself is ( ) N m C E (0.05 m) ( C)ŷ N/C ŷ or 10 kn/c 3

4 Problems Both of these problems require ou to remember a bit from Phsics A uniform electric field (that is, E is the same at ever point) eists in a region between oppositel charged plates. An electron is released from rest at the surface of the negativel charged plate and strikes the surface of the opposite plate,.0 cm awa, in a time s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field E? There is a uniform electric field between the plates, so the electron eperiences a constant force, and thus undergoes constant acceleration. Therefore we can use all of those wonderful constant-acceleration formulae we learned in mechanics when dealing with gravit. Specificall, 0 + v 0 t + 1 at and v v 0 + at Solving the second equation for a gives us a (v v 0 )/t; substituting into the first equation gives us 0 v 0 t + 1 (v v 0)t 1 (v + v 0)t which is a less familiar constant-acceleration formula (at least to me): the distance travelled is equal to the time travelled times the average speed, which in the case of constant acceleration is (v + v 0 )/. The initial velocit of the electron is v 0 0, so the speed of the electron as it strikes the second plate is v t (0.0 m) s m/s which answers part a. For part b, we need the electric field, which means we need the force, which means we need the acceleration: E F q ma q The acceleration comes from the equation v v 0 + at and so a v t m/s s m/s. The charge of an electron is q C and the mass of an electron is m kg, so E ( kg)( m/s ) C 1000 N/C. 5. In the figure, a small, nonconducting ball of mass m 1.0 mg (note the units) and charge q C (distributed uniforml through its volume) hangs from an insulating thread that makes an angle θ 30 with a vertical sheet which has a uniform charge densit σ (shown in cross section); such a sheet creates a uniform electric field which points awa from or towards the sheet (depending on the sign), with magnitude E σ. Considering the gravitational force on the ɛ 0 ball and assuming the sheet etends far verticall and into and out of the page, calculate the surface charge densit σ of the sheet. 4

5 There are three forces on the ball, as shown in the figure to the right: the tension T in the string, the weight mg of the ball, and the electric force qe due to the field of the plate. The tension force vector can be broken into its components T cos θ pointing upward and T sin θ pointing to the left. Assuming the ball is not moving, the net force on the ball is zero, and so T sin θ qe and T cos θ mg We use the second equation to find T : T mg cos θ ( kg)(9.8 m/s ) cos N N 3/ Using this value, we solve for the electric field at the location of the ball: E T sin θ q ( N) sin C 83 N/C Now the electric field due to an infinite sheet of charge is E πkσ. Therefore, σ 1 πk E 1 π( ) (83 N/C) C/m 5.00 nc/m. 5