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Idea of Symmetry

An electrosta...

Question

An electrostatic field line leaves at an angle $α$ from point charge $q_{1}$ , and connects with point charge $- q_{2}$ at an angle $β$ , as shown schematically in the figure. We have $| q_{1} | = | 2 q_{2} |$ and $α = 60^{\circ}$ . Then:

A test charge placed at the midpoint

O

will be in equilibrium

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β = 90^{\circ}

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The magnitude of the electric field at a distance

r

far away from

O

q_{2} / 4 π ϵ_{0} r^{2}

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For a negative test charge, the electrostatic potential energy is lower near

q_{1}

as compared to that near

q_{2}

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Solution

The correct options are
C The magnitude of the electric field at a distance $r$ far away from $O$ is $q_{2} / 4 π ϵ_{0} r^{2}$
D For a negative test charge, the electrostatic potential energy is lower near $q_{1}$ as compared to that near $q_{2}$
In this projection the magnitude of electric field is same as the magnitude of electric field.
Now, the magnitude of the electric field at a distance $r$ far away from $0$ is $\frac{q_{2}}{4 π ϵ_{0} r^{2}}$ and for a negative test charge, the electrostatic potential energy is lower near $q_{1}$ as compared to that near $q_{2}$
When the negative test charge is coming towards $+ q_{1}$ and $+ q_{2}$ the repulsion occurs.

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An electrostatic field line leaves at an angle α from point charge q1, and connects with point charge −q2 at an angle β, as shown schematically in the figure. We have |q1|=|2q2| and α=60∘. Then:

An electrostatic field line leaves at an angle $α$ from point charge $q_{1}$ , and connects with point charge $- q_{2}$ at an angle $β$ , as shown schematically in the figure. We have $| q_{1} | = | 2 q_{2} |$ and $α = 60^{\circ}$ . Then: