The vectors A- and D- in the figure have the following magnitudes: A=8.05 m and

Question

The vectors A- and D- in the figure have the following magnitudes: A=8.05 m and D = 9.6 m(Assume this is a right-handed coordinate system.) Find the magnitude and direction of A- times D Find the magnitude and direction of D- times A-. In a rectangular coordinate system a positive point charge q = 7.20 times 10^-9 C is placed at the point x = +0.150 m, y = 0, and an identical charge is placed at x = -0.150 m, y = 0. Find the x- and y components, the magnitude of the electric field at the following points. A proton is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v_o and is directed at an angle a below the horizontal. find the maximum distance h_max that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (Use any variable or symbol stated above as necessary.) After what horizontal distance d does the proton return to its original elevation? (Use any variable or symbol stated above as necessary.) Sketch the trajectory of the proton. Find the numerical values of h_max and d if E = 480 N/C, v_0 = 4.50 times 10^5 m/s and a = 31.0 degree.

Explanation / Answer

a) at (0,0)

the field due to two charges are same and opposite in direction ,hence they will cancel

so the fied at (0,0) is 0 N/C

b) at (0.3,0)

the field due to charge at (0.15,0) is E1 = k*q/r1^2 = (9*10^9*7.2*10^-9)/(0.3-0.15)^2 = 2880 N/C

the field due to charge at (-0.15,0) is E2 = k*q/r2^2 = (9*10^9*7.2*10^-9)/(0.3+0.15)^2 = 320 N/C

So net field is E = E1+E2 = 2880+320 = 3200 N/C

Enet,x = 3200 N/C

Enet,y = 0 N/C
Enet = 3200 N/C

c) at (0.15,-0.4)

E1 = k*q/r1^2 = (9*10^9*7.2*10^-9)/(0.4^2) = 405 N/C along -Y direction

E2 = k*q/r2^2 = (9*10^9*7.2*10^-9)/(0.3^2+0.4^2) = 259.2 N/C

X-component of E2 is E2x = E2*cos(theta) = 259.2*(0.3/sqrt(0.3^2+0.4^2)) = 155.52 N along +X-axis

y-component of E2 is E2y = E2*sin(theta) = 259.2*(0.4/sqrt(0.3^2+0.4^2)) = 207.36 along-Y-axis

then x-component of net field is Enet,x= 155.52 N/c

y-component of net field is Enet,y = 405+207.36 = 612.36 N/C along -Y-axis

then Enet = sqrt(155.52^2+612.36^2) = 631.8 N/C

D) X-components due to two charges cancels out and hence the Enet,x = 0 N/C

Enet,y = E1y+E2y = (k*q/r^2)*(0.2/sqrt(0.15^2+0.2^2)) = (9*10^9*7.2*10^-9)*(0.2/sqrt(0.15^2+0.2^2)) = 51.84 N/C

Enet = 51.84 N/C

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