Suppose the particle is placed at the 40V equipotential surface along the line connecting the two central positive and negative charges. (Hint: The electric field can be obtained using the 40V and 30V potential difference lines.)

1. What is the force on the charged particle if q=80μC? (mN) (m-milli 10-3)

2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN)

Hello some part of your question is missing below is the missing part

2.  What is the force on the charged particle if it is now located at the 0V potential difference line? (mN) (hint: The electric field can be obtained as above using the 0V and -10V equipotential lines.)

answer :

1) 0.8 mN

2) 0.8 mN

Explanation:

Given data:

1) Calculate the force on the charged particle

q = 80 μC , Va = 30v , Vb = 40v, ∝ = 1 m

E = ( Δv ) / ∝

   = ( Vb - Va ) / ∝

F = qE

  = 80 μC * ( 40 - 30 ) / 1 m

  = 800 μC

F = 0.8 mN

2) Calculate the force on the charged particle when it is located at 0V

Va = -10V , Vb = 0V, q = 80 μC,  ∝ = 1 m

F = qE

   where E = ( 0 - ( -10 ) / 1  

F = 80 μC *  ( 0 - ( -10 ) / 1

  = 800 μC = 0.8 mN

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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