Two equal and opposite charges of magnitude 2 x10⁻⁷ C are placed 15 cm apart. What is the magnitude and direction of electric intensity (E) at a point midway between the charges? What force would act on a proton (charge = +1.6 x 10⁻¹⁹ C) placed there? (Answer 0.64 x 10⁶ N/C along AB, 1.024 x 10 ⁻¹³ N along AB)   



Given:

Let 

1st charge = q₁ = 2 x 10⁻⁷ C 

2nd charge = q₂ = - 2 x 10⁻⁷ C 

Separation distance
 = r = 15 cm = 0.15 m

Midway point 'P' distance from each charges = d = d₂ = d =  r/2 = 0.15 m/2 = 0.075 m

Charge on Proton = qₚ = +1.6 x 10⁻¹⁹ C

∴ Coulomb's Constant = K = 14ε0 = 9 x 10⁹ Nm²C⁻²

To Find:

The magnitude of the Electric Field at midway point P  = |Eₙₑₜ| = ?

The magnitude of the force at the midway point on proton  = Fp = ?


Solution:

The diagram can be drawn according to the given conditions are  as below



The direction of the electric field at point P due to both charges q₁(+ve) and q₂(-ve) are in the same direction along AB and the magnitude is given by:

E = Eq +Eq

E  = kqd2 ˆd + kqd2 ˆd


As q₂ = -q₁ {opposite and equal charges} , d₁ = d₂ = d and ˆd = - ˆd {opposite in direction} so,


E  = kqd2 ˆd + k-qd2 (-ˆd)


E  = kqd2 ˆd + kqd2 ˆd


E  = 2 kqd2 ˆd


E = 2 kqd2 ˆd


E = 2 kqd2ˆd


by putting values


E = 2 x 9 x 10⁹ Nm²C⁻² x 210C(0.075m)2 ˆd


E = 18 x 10⁹ Nm²C⁻²210C0.005625m² ˆd


E= 18 x 10⁹ Nm²C⁻² x 355.555x 10⁻⁷ C m⁻² ˆd


E = 6400 x 10⁹⁻⁷ ˆd NC⁻¹ 


E = 0.64 x 10⁴ x 10² ˆd NC⁻¹


E  = 0.64 x 10⁶  ˆd NC⁻¹

Taking absolute value and  |ˆd| =1 

|Eₙₑₜ| = 0.64 x 10⁶ NC⁻¹ and will be along AB direction ---------------Ans (1)


Now to find Fp

Fp = E x qₚ

by putting values

Fp = 0.64 x 10⁶  ˆd NC⁻¹ x 1.6 x 10⁻¹⁹ C

Fp = 1.024 x 10⁻¹³ N ----------------Ans.(2)

It is along the line of AB



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