A point charge q = +0.87 nC is fixed at the origin. Where must a proton be place

Question

A point charge q = +0.87 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on the proton to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)

? m

An object of mass m = 2.0 g and charge Q = +46 µC is attached to a string and placed in a uniform electric field that is inclined at an angle of 30.0° with the horizontal. The object is in static equilibrium when the string is horizontal.

(a) Find the magnitude of the electric field.
N/C
(b) Find the tension in the string.
N

Explanation / Answer

We are looking for in the table the mass and charge of the proton

. mp = 1.672 10-27 Kg

. q = 1.6 10-19 C

Usamos las leyes de Newton y coulomb

Fe – W = 0     Fe = W                   k Q q/ r2= m g

. r2 = k Q q/ mg         r2 = 9 109   0.87 10-9 1.6 10-19 / 1.672 10-27 9.8

.   r2 = 6.23 /16.38    108 m = 0.3803   108 m2

.   r = 0.6167 104     m

We place the positive charge of the proton on the shaft and directly below the origin

Second Part

The only inclined force is the electric force

Fe = Fex i + Fey j      Fex= Fe Cos                      Fey = Fe sin

We use Newton's second law Axis X

Fex – T = 0

Axis Y    

Fey – W =0          Fe Sin = m g           E q Sin = mg             

E = mg/ q Sin                     E = 2 10-3 9.8 / (46 10-6 Sin 30) = 852.17 N/C

E= 852.17 N/C

Fex = T                                Fe Cos = T          T= E q Cos      T= 852.17 46 10-6 Cos 30

T= 0.33948 N

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