Three point charges q_1 = +2.0 \times 10^{-6}\text{ C}, q_2 = +2.0 \times 10^{-6}\text{ C}… — JAMB UTME

Three point charges $q_1 = +2.0 \times 10^{-6}\text{ C}$ , $q_2 = +2.0 \times 10^{-6}\text{ C}$ , and $q_3 = -4.0 \times 10^{-6}\text{ C}$ are placed along a straight line at positions $x = 0\text{ m}$ , $x = 0.30\text{ m}$ , and $x = 0.60\text{ m}$ , respectively. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , calculate the magnitude of the net electrostatic force acting on charge $q_2$ in newtons.

Cevap: 1.2 N

Cevap

The magnitude of the net electrostatic force acting on charge

q_2

1.20\text{ N}

The force exerted on

q_2

q_1

is repulsive (

0.40\text{ N}

directed to the right) because both charges are positive. The force exerted on

q_2

q_3

is attractive (

0.80\text{ N}

directed to the right) because

q_2

is positive and

q_3

is negative. Since both component forces act in the same direction, the total net force magnitude is

0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Adım Adım Çözüm

Calculate the repulsive force exerted by

q_1

q_2

F_{12} = 0.40\text{ N}

pointing to the right

Like charges repel each other, so

q_1

pushes

q_2

away along the

+x

-axis.

Calculate the attractive force exerted by

q_3

q_2

F_{32} = 0.80\text{ N}

pointing to the right

Unlike charges attract each other, so

q_3

pulls

q_2

towards itself along the

+x

-axis.

Sum the component electrostatic forces acting on

q_2

F_{\text{net}} = 0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Because both forces act in the exact same direction along the line, their magnitudes add directly.

Anahtar Kavram

Coulomb's Law and Principle of Superposition for Electrostatic Forces

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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