Question

In: Physics

1. a uniform line of charge with total chRge Q runs along the y axis from...

1. a uniform line of charge with total chRge Q runs along the y axis from -l to 0. What is the electric field vector at the position (-x,0)? Leave your answer in the form of two well defined integrals. (one for x component one for y compenent)

2. a 25 microcoulumb charge is located at (2,0), a -170 microcoulumb charge is loacted at (0,5), and a 100 microcoulumb charge is located at the origin. ALL positions are given in CENTIMETERS.

what is the magnitude of tbe net force acting on the 25 microcoulumb charge?

what is the unit vector that points in the same direction as the electric field at the point (2,5)?

3. a proton starts at position x=3 cm and ana electron starts at p= -3cm. what best describes the region where they will collide?

x<-3. -3<x<0. 0<x<3 3<x or they will not collide

Expert Solution

Consider the Electric field due to a line distribution of charge to be dE

Charge (total)=Q

Charge per unit length =Q/(0-(-l))=Q/l=

E at distance x on x-axis=9×10⁹×Q/l×dy/r²

Where r=

dE=9×10⁹Q/l×dy/(x²+y²)

Integrating both sides,

E=9×10⁹Q/ldy

E=9×10⁹×Q/l×dy

E=9×10⁹×Q/l×tan^-1_(l/x)/x

2.Consider force on charge located at (2,0) because of 100 microcoulomb charge which is given by coulombs law as

F1=9×10⁹×q1q2/r²=9×10⁹×25×10^-10/(2²×10^-4)=5.625kN

Similarly Force on charge 25 micro coulomb by charge -170 microC(F2)=-170×10^-6×25×^-6×9×10⁹/((5²+2²)×10^-4)=-13.189kN

Thus net force can obtained by principle of superposition of F1 and F2 as:

F=F1+F2=(-13.189+5.625)kN=7.564kN

2.Electric field at point (2,5) due to superposition of fields

=9×10^9×10^-6((-1700000/4)+(1000000/29)+(250000/25))

=3825+310.34)+90In terms of vector notation

=4135.34i +400.34j kN

Value or magnitude=sqrt(4135²+400²)

=4154.64kN

Therefore approx unit electric field vecto=4135.34/4154.6i+400.34÷4154.6j =0.995i+0.096jAns.

3.consider the proton and electron collide at xcm

We know that according to the relation

ma=-k.q1q2/r²

(equivalence of mechanical and electrostatic forces =》mvdv/dx=kq1q2/r²

=>1/2mv²=k.q1q2/r the negative sign in first equation is due to opposite sign of proton and electron.)

We can say distance of collision r is inversely proportional to mass m.

Thus the particles collides (since masses of proton>>>mass of electron.

near to proton and away from electron.

Thus point of collision is (0,3)=》x belongs ti 0<x<3 cm.

genius_generous answered 2 hours ago

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