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Stells [14]
3 years ago
8

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

e magnitude of the electric field at the origin?
Physics
1 answer:
IceJOKER [234]3 years ago
8 0

Answer:E=75\ N/m

Explanation:

Given

First charge of q_1=15\ nC is placed at x=1.5\ m

Second charge  q_2=-20\ nC is placed at y=-2\ m

Electric field is given by

E=\frac{kq}{r^2}

Electric field due to q_1 is away from it

E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}

E_1=60\ N/m

Electric field due to q_2

E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}

E_2=45\ N/m

Net electric field will be vector addition of two

\vec{E_{net}}=\vec{E_1}+\vec{E_2}

\vec{E_{net}}=-60\hat{i}-45\hat{j}

Magnitude of Electric field is

E=\sqrt{60^2+45^2}

E=75\ N/m

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Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

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r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

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taking square root on both sides:

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Incomplete question. Missing part:

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