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3 years ago

7

What is a fundamental difference between gravitational forces and electrostatic forces?

1 answer:

Kryger [21]3 years ago

8 0

Answer:

The differences are: Electrostatic force is force between two charges separated by distance in space while gravitational force is the force between two masses separated by a distance. Electrostatic force is either repulsive or attractive while gravitational force is always attractive.

Explanation:

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A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗

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The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

k is coulomb's constant
q₁ and q₂ are the charges
r is the distance between the charges

<h3>Distance between the charges</h3>

$|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units$

$F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \ N$

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

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2 years ago

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

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