Five charges, each +Q, are placed at five ver | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In a regular hexagon, the distance from the center to any vertex is equal to the side length $L$. Let the vertices be $V_1, V_2, V_3, V_4, V_5, V_

Vedclass · Accepted Answer

$k\frac{Q^2}{L^2}$

Vedclass · Answer

(A) In a regular hexagon, the distance from the center to any vertex is equal to the side length $L$. Let the vertices be $V_1, V_2, V_3, V_4, V_5, V_6$. Suppose charges $+Q$ are placed at $V_1, V_2, V_3, V_4, V_5$ and the center $O$ has a charge $-Q$.If there were a charge $+Q$ at $V_6$ as well, the net force on the center $O$ would be zero due to symmetry (as each pair of opposite charges would exert equal and opposite forces).Let $\vec{F}_1, \vec{F}_2, \vec{F}_3, \vec{F}_4, \vec{F}_5$ be the forces exerted by the charges at the five vertices on the center charge $-Q$. Let $\vec{F}_6$ be the force that would be exerted by a charge $+Q$ at $V_6$.By symmetry, $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 + \vec{F}_6 = 0$.Therefore, the net force $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 = -\vec{F}_6$.The force $\vec{F}_6$ exerted by a charge $+Q$ at $V_6$ on $-Q$ at the center is attractive and has a magnitude $F = k\frac{|(+Q)(-Q)|}{L^2} = k\frac{Q^2}{L^2}$.Since $\vec{F}_{net} = -\vec{F}_6$, the magnitude of the net force is $|\vec{F}_{net}| = |-\vec{F}_6| = k\frac{Q^2}{L^2}$.

Five charges, each $+Q$, are placed at five vertices of a regular hexagon of side length $L$. What is the magnitude of the force on a charge $-Q$ placed at the center of the hexagon?

Similar Questions

Three equal charges are placed on the three corners of a square as shown below. If the magnitude of force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,then the ratio of $F_{13}$ to $F_{12}$ is

When two identical point charges are placed at a distance of $5 \, cm$, they experience a repulsive force of $0.144 \, N$. The value of each charge in microcoulombs $(\mu C)$ is:

For the system shown in the figure,find $Q$ such that the resultant force on $q$ is zero.

In a hydrogen atom,the distance between the electron and proton is $2.5 \times 10^{-11} \ m$. The electrical force of attraction between them will be

Two point charges of $+2 \, \mu \text{C}$ and $+6 \, \mu \text{C}$ repel each other with a force of $12 \, \text{N}$. If a charge of $-4 \, \mu \text{C}$ is added to each,the force will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module