Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2022

PhysicsEasysingle choice

The three charges $\frac{q}{2}, q$ and $\frac{q}{2}$ are placed at the corners $A, D$ and $C$ of a square of side $a$ as shown in figure. The magnitude of electric field $(E)$ at the corner $B$ of the square, is

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Answer:

Solution:

$E_{A} = \frac{k (\frac{q}{2})}{a^{2}} = \frac{k q}{2 a^{2}}$ , $E_{C} = \frac{k (\frac{q}{2})}{a^{2}} = \frac{k q}{2 a^{2}}$ and $E_{D} = \frac{k (q)}{{(\sqrt{2} a)}^{2}} = \frac{k q}{2 a^{2}}$

As we can see from diagram, $E_{C} & E_{A}$ are perpendicular.

Therefore, net electric field will be,

$= \frac{k q}{2 a^{2}} + \sqrt{{(\frac{k q}{2 a^{2}})}^{2} + {(\frac{k q}{2 a^{2}})}^{2}} = \frac{k q}{2 a^{2}} + \frac{k q}{\sqrt{2} a^{2}} = \frac{q}{4 π ϵ_{0} a^{2}} (\frac{1}{\sqrt{2}} + \frac{1}{2})$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: PYQ_2022

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