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Sagot :
The charge on the particle when the 4000 v equipotential surface is 26. 0 cm farther from a positively charged particle than the 5000 v equipotential surface is 2.89×10⁻⁵ nC.
How to find the point charge due?
The point charge can be found using the following expression.
[tex]q=\dfrac{Vr}{K}[/tex]
Here, V is the potential of a point charge, r is the distance and K is the coulombs constant.
This equation can also be given as,
[tex]r=K\dfrac{q}{V}[/tex]
The 4000 V equipotential surface is 26.0 cm farther from a positively charged particle than the 5000 v equipotential surface. For this case, the distance at both point is,
[tex]r_1=K\dfrac{q}{V_1}\\r_2=K\dfrac{q}{V_2}[/tex]
Subtract both the distance which is equal to 26 cm.
[tex]r_1-r_2=K\dfrac{q}{V_1}-K\dfrac{q}{V_2}\\26=Kq(\dfrac{1}{V_1}-\dfrac{1}{V_2})\\26=(9\times10^9)q(\dfrac{1}{4000}-\dfrac{1}{5000})\\q=2.89\times10^{-14}\rm\; C\\q=2.89\times10^{-5}\rm\; nC[/tex]
Thus, the charge on the particle when the 4000 v equipotential surface is 26. 0 cm farther from a positively charged particle than the 5000 v equipotential surface is 2.89×10⁻⁵ nC.
Learn more about the point charge here;
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