Menu Close

Problem 4: Two opposite charges of magnitude 2 × 10-7 are placed 15 cm apart. What is the magnitude and direction of electric intensity (E) at a point mid-way between the charges? What force would act on a proton placed there?


Consider the figure. Since both the charges are opposite, therefore, electric intensity at any point between the charges would have the same direction; likewise, the midpoint.

We know that the electric intensity at a certain point p due to a point charge q is given by,

So the resultant electric field intensity due to both charges is,

Since the distance between the charges is 15 cm and the point is mid-way between them, therefore, its distance from both charges is 15/2 = 7.5 cm = 0.075 m. Put r = 0.075 m,

Now to find the force on a proton placed at this point, we have, F = Eq

Put the values,


  1. Pingback:numerical-problem-5-chapter-1-electrostatics-physics-12-msa – msa

  2. Pingback:numerical-problems-on-electrostatics-chapter-1-physics-12-msa – msa

  3. Pingback:numerical-problem-3-chapter-1-electrostatics-physics-12 – msa

Leave a Reply

Your email address will not be published. Required fields are marked *