Question

Please answer the question

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{E_{net}=1.8\times 10^{8}\:N/C}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt:\implies Two \: charges = + 4 \mu C \: and \: + 6 \mu C \\ \\ \tt :\implies Distance \: between \: charges = 2 \: cm \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Net \: electric \: field \: at \: mid \: point = ?$

• According to given question :

$\tt \circ \: q_{1} = 4 \mu \: C =4\times 10^{-6}\:C\\ \\ \tt \circ \:r = 1 \: cm = 0.01 \: m\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies E_{1} = \frac{k q_{1} }{ {r}^{2} } \\ \\ \tt: \implies E_{1} = \frac{k \times 4\times 10^{-6}}{ {(0.01)}^{2} } \\ \\ \tt: \implies E_{1} =0.04k \: N/C- - - - - (1) \\ \\ \tt \circ \: q_{2} = 6 \mu \: C =6\times 10^{-6}\:C \\ \\ \tt \circ \: r = 1 \: cm = 0.01 \: m\\ \\ \bold{Similarly : } \\ \tt: \implies E_{2} = \frac{k q_{2} }{ {r}^{2} } \\ \\ \tt: \implies E_{2} = \frac{k \times 6\times 10^{-6}}{ {(0.01)}^{2} } \\ \\ \tt: \implies E_{2} = 0.06k \: N/C- - - - - (2) \\ \\ \bold{For \: net \: electric \: field: } \\ \tt: \implies E_{net} = E_{2} - E_{1} \\ \\ \tt: \implies E_{net} = 0.06k - 0.04k \\ \\ \tt: \implies E_{net} = 0.02k \: \\ \\ \tt: \implies E_{net} = 0.02 \times 9 \times {10}^{9} \\ \\ \green{\tt: \implies E_{net} =1.8 \times {10}^{8} \: N/C}$

Answer 2

GIVEN:

• Two charges +4μc & +6μc
• Distance between the charges = 2cm

TO FIND:

• Net electric field intensity at mid point

SOLUTION:

As per the question

q₁ = 4μc = 4 × 10-⁶

q₂ = 6μc = 6 × 10-⁶

r = 1cm = 0.01 m = 10-² m

__________________

We know that

→ E₁ = kq₁/r²

→ E₁ = k × 4 × 10-⁶/(10-²)²

→ E₁ = 0.04kN/C

__________________

Similarly

Finding E₂

→ E₂ = kq₂/r²

→ E₂ = k × 6 × 10-⁶/(10-²)²

→ E₂ = 0.06kN/C

Finding net electric field (Eₙₑₜ)

→ Eₙₑₜ = E₂ - E₁

→ Eₙₑₜ = 0.06 - 0.04

→ Eₙₑₜ = 0.02kN/C ……( Answer )