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:
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 = 14ㄫε0 = 9 x 10⁹ Nm²C⁻²
To Find:
The magnitude of the Electric Field at midway point P = |Eₙₑₜ| = ?
→Eₙₑₜ = Eq₁ +Eq₂
→Eₙₑₜ = kq₁d₁2 ˆd₁ + kq₂d₂2 ˆd₂
As q₂ = -q₁ {opposite and equal charges} , d₁ = d₂ = d and ˆd₂ = - ˆd₁ {opposite in direction} so,
→Eₙₑₜ = kq₁d2 ˆd₁ + k-q₁d2 (-ˆd₁)
→Eₙₑₜ = kq₁d2 ˆd₁ + kq₁d2 ˆd₁
→Eₙₑₜ = 2 kq₁d2 ˆd₁
→Eₙₑₜ = 2 kq₁d2 ˆd₁
→Eₙₑₜ = 2 kq₁d2ˆd₁
by putting values
→Eₙₑₜ = 2 x 9 x 10⁹ Nm²C⁻² x 2x10⁻⁷C(0.075m)2 ˆd₁
→Eₙₑₜ = 18 x 10⁹ Nm²C⁻²2x10⁻⁷C0.005625m² ˆd₁
→Eₙₑₜ= 18 x 10⁹ Nm²C⁻² x 355.555x 10⁻⁷ C m⁻² ˆd₁
→Eₙₑₜ = 6400 x 10⁹⁻⁷ ˆd₁ NC⁻¹
→Eₙₑₜ = 0.64 x 10⁴ x 10² ˆd₁ NC⁻¹
→Eₙₑₜ = 0.64 x 10⁶ ˆd₁ NC⁻¹
Taking absolute value and |ˆd₁| =1
|Eₙₑₜ| = 0.64 x 10⁶ NC⁻¹ and will be along AB direction ---------------Ans (1)
Now to find →Fp
→Fp = →Eₙₑₜ x qₚ
by putting values
→Fp = 0.64 x 10⁶ ˆd₁ 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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