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 = d₁ = 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 = `\frac {1}{4ㄫε_0}` = 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  = `\vec F_{p}` = ?


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:

`\vec Eₙₑₜ` = `E_{q₁}` +`E_{q₂}`

`\vec Eₙₑₜ`  = k`\frac {q₁}{d₁^2}` `\hat d₁` + k`\frac {q₂}{d₂^2}` `\hat d₂`


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


`\vec Eₙₑₜ`  = k`\frac {q₁}{d^2}` `\hat d₁` + k`\frac {-q₁}{d^2}` (-`\hat d₁`)


`\vec Eₙₑₜ`  = k`\frac {q₁}{d^2}` `\hat d₁` + k`\frac {q₁}{d^2}` `\hat d₁`


`\vec Eₙₑₜ`  = 2 k`\frac {q₁}{d^2}` `\hat d₁`


`\vec Eₙₑₜ` = 2 k`\frac {q₁}{d^2}` `\hat d₁`


`\vec Eₙₑₜ` = 2 k`\frac {q₁}{d^2}``\hat d₁`


by putting values


`\vec Eₙₑₜ` = 2 ｘ 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {2 ｘ 10⁻⁷ C }{(0.075 m)^2}` `\hat d₁`


`\vec Eₙₑₜ` = 18 ｘ 10⁹ Nm²C⁻²`\frac {2 ｘ 10⁻⁷ C }{0.005625 m²}` `\hat d₁`


`\vec Eₙₑₜ`= 18 ｘ 10⁹ Nm²C⁻² ｘ 355.555ｘ 10⁻⁷ C m⁻² `\hat d₁`


`\vec Eₙₑₜ` = 6400 ｘ 10⁹⁻⁷ `\hat d₁` NC⁻¹ 


`\vec Eₙₑₜ` = 0.64 ｘ 10⁴ ｘ 10² `\hat d₁` NC⁻¹


`\vec Eₙₑₜ`  = 0.64 ｘ 10⁶  `\hat d₁` NC⁻¹

Taking absolute value and  |`\hat d₁`| =1 

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


Now to find `\vec F_p`

`\vec F_p` = `\vec Eₙₑₜ` ｘ qₚ

by putting values

`\vec F_p` = 0.64 ｘ 10⁶  `\hat d₁` NC⁻¹ ｘ 1.6 ｘ 10⁻¹⁹ C

`\vec F_p` = 1.024 ｘ 10⁻¹³ N ----------------Ans.(2)

It is along the line of AB



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