Two charges 2C and -2C are at 0.01m. Calculate the electric field intensity 0.3m in equatorial plane.
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Expert's answer
2021-06-07T08:30:35-0400
Let us consider two equal and opposite charges q and −q at points A and B separated by a distance 2d as shown in the figure. We have to find the electric field at an equatorial point P distant x from the center O .
From ΔAOP and ΔBOP ,
AP=BP=x2+d2
The electric field at P due to q is
Ea=4πϵ01x2+d2q directed along AP.
The electric field at P due to −q is
Eb=4πϵ01x2+d2q directed along PB.
Note that the electric fields Ea and Eb have equal magnitude i.e,
∣Ea∣=∣Eb∣=E(say)
To find the resultant electric field at P we will resolve Ea and Eb in horizontal and vertical components as shown below.
The vertical components Easinθ and Ebsinθ will cancel each other (since ∣Ea∣=∣Eb∣ ).
The horizontal components Eacosθ and Ebcosθ will add up to give the resultant electric field.
∴ Resultant electric field at P is
Er=Eacosθ+Ebcosθ⇒Er=2Ecosθ⇒Er=24πϵ01x2+d2q.x2+d2d⇒Er=4πϵ01(x2+d2)3/2q.2d directed along AB.
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