Question

Two charges are placed on the x axis. One of the charges (ai +9.0 pC) s atx3.2 cm and the other ( 22 pC) s at x8.3 om )Find the net electric field (magnitude and direction) at x-0 cm. (Use the sign of your answer to Indicate the direction along the x-axis.) (b) Find the net electric fleld (magnitude and direction) at x5.9 cm. (Use the sign of your answer to indicate the direction along the x axls.) N/C Secion 181 electric fele thot is directed aiong the axs (see the foure) The proton starts with a velocity of 120 10 m/s In a vacuum, a proton (charge e, mass 1.67 10" kg) is moving paralel to a uniform and accelerates in the same direction as the electric held, which has a velu, of ,2.1×10 Ne hdthe velo ity ofthe pro direction with the sign of your answer hnits dis s met i,'+2 from the en ing m/s Additional Materials

Answer 1

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2.

In a Vacuum, a proton.......

Ans:

Relation between force and electric field is given by:

F = q*E

Since q = charge on proton (which is +ve) = +e = +1.6*10^-19 C

F = e*E

Since Electric field is towards right and charge in field is +ve, So force on proton will be in the same direction of electric field, which is towards right.

Now Using Force balance, From newton's 2nd law:

Fnet = m*a

So,

m*a = q*E

a = q*E/m

Now Using given values, acceleration will be

m = mass of proton = 1.67*10^-27 kg

a = 1.6*10^-19*2.1*10^3/(1.67*10^-27)

a = 2.01*10^11 m/sec^2

Vi = 2.0*10^4 m/sec

S = distance traveled = 3.0 mm = 3.0*10^-3 m

Using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*a*S

Vf = sqrt ((2.0*10^4)^2 + 2*(2.01*10^11)*(3.0*10^-3))

Vf = 40074.9 m/sec

Vf = 4.01*10^4 m/sec

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