Two charges, Q1= 3.90 ?C, and Q2= 6.10 ?C are located at points (0,-2.50 cm ) an

Question

Two charges, Q1= 3.90 ?C, and Q2= 6.10 ?C are located at points (0,-2.50 cm ) and (0,+2.50 cm), as shown in the figure.

1. What is the y-component of the total electric field at P?

2. What is the magnitude of the total electric field at P?

3. Now let Q2 = Q1 = 3.90 ?C. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

4.Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

Lets asume

Distance Between origine and Point P is x ,

Distance Between Q1/Q2( any ) and Point P is r,

1) Y component of field ,   

Ey = k*Q1/r^2*sin( x/r ) + k*Q2/r^2*sin( x/r )

Ey = 9*109 X (3.90 C )/r^2*sin( x/r ) + 9*109 X *(6.10 C)/r^2*sin( x/r )

2)

mag = sqrt(Ex^2 + Ey^2)

= sqrt [ {  9*109 X (3.90 C )/r^2*cos( x/r ) + 9*109 X *(6.10 C)/r^2*cos( x/r ) }/2 + { 9*109 X (3.90 C )/r^2*sin( x/r ) + 9*109 X *(6.10 C)/r^2*sin( x/r ) } /2 ]

3 ) X component is ZEro ,, so Y component will be

2*9*109 X (3.90 C ) /r^2*sin( x/r  )

4 ) ) F = E*q = 2*9*109 X (3.90 C )   /r^2*sin(  x/r  ) X 1.60x10^-19C

( where 1.60x10^-19C is charge of electron )

here i have wriiten  C at functions to avoid length of equation to convert it in C you need to multiply this value with 10-6

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