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Electric Charges and Fields - Forces between Multiple Charges

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Principle of Superposition: When multiple charges are present, the total force on any one charge is the vector sum of the forces exerted on it by all other charges, taken one at a time.

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Independence of Forces: The individual force between any two charges is unaffected by the presence of other charges.

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Vector Addition: Since electrostatic force is a vector quantity, the net force is calculated using the parallelogram law of vector addition or the method of resolution of components.

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Geometry and Symmetry: For symmetric charge distributions (like charges at vertices of a regular polygon), the net force at the geometric center is often zero if the charges are identical.

📐Formulae

F⃗1=F⃗12+F⃗13+⋯+F⃗1n\vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + \dots + \vec{F}_{1n}

F⃗1=14πϵ0∑j=2nq1qjr1j2r^1j\vec{F}_{1} = \frac{1}{4\pi\epsilon_0} \sum_{j=2}^{n} \frac{q_1 q_j}{r_{1j}^2} \hat{r}_{1j}

Fnet=F12+F22+2F1F2cos⁡θF_{net} = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos \theta}

tan⁡β=F2sin⁡θF1+F2cos⁡θ\tan \beta = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}

💡Examples

Problem 1:

Three point charges qq, qq, and −q-q are placed at the vertices of an equilateral triangle of side ll. Find the magnitude of the net force acting on the charge −q-q.

Solution:

Let the charges at vertices AA and BB be +q+q and the charge at vertex CC be −q-q. The force exerted by AA on CC is FCA=14πϵ0q2l2F_{CA} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2} (attractive). The force exerted by BB on CC is FCB=14πϵ0q2l2F_{CB} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2} (attractive). The angle between these two forces is 60∘60^\circ. Using the resultant formula: Fnet=F2+F2+2F2cos⁡60∘F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cos 60^\circ} Fnet=2F2+2F2(12)=3F2=F3F_{net} = \sqrt{2F^2 + 2F^2 (\frac{1}{2})} = \sqrt{3F^2} = F\sqrt{3} Substituting FF, we get Fnet=314πϵ0q2l2F_{net} = \sqrt{3} \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2}.

Explanation:

Since the forces are vectors, we cannot simply add their magnitudes. Because both forces are directed towards the other vertices along the sides of the equilateral triangle, the angle between the force vectors is the interior angle of the triangle (60∘60^\circ).

Problem 2:

Two point charges q1=+4μCq_1 = +4\mu C and q2=+1μCq_2 = +1\mu C are separated by a distance of 30 cm30\text{ cm}. Where should a third charge qq be placed on the line joining them so that it experiences no net force?

Solution:

Let the charge qq be placed at a distance xx from q1q_1. Then its distance from q2q_2 is (30−x)(30 - x). For equilibrium: 14πϵ0q1qx2=14πϵ0q2q(30−x)2\frac{1}{4\pi\epsilon_0} \frac{q_1 q}{x^2} = \frac{1}{4\pi\epsilon_0} \frac{q_2 q}{(30-x)^2} 4x2=1(30−x)2\frac{4}{x^2} = \frac{1}{(30-x)^2} Taking the square root on both sides: 2x=130−x\frac{2}{x} = \frac{1}{30-x} 60−2x=x  ⟹  3x=60  ⟹  x=20 cm60 - 2x = x \implies 3x = 60 \implies x = 20\text{ cm}

Explanation:

The net force is zero when the magnitudes of the forces exerted by q1q_1 and q2q_2 are equal and their directions are opposite. This occurs between two like charges.