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Question: Two charges \[ + {1.10^{ - 6}}\] and \[ - {4.10^{ - 6}}\] are separated by a distance of \[2\,m\]. W...

Two charges +1.10−6 + {1.10^{ - 6}} and −4.10−6 - {4.10^{ - 6}} are separated by a distance of 2 m2\,m. Where is the null point located ?

Explanation

Solution

In the above question we were asked to find null points between two charges placed with distance between. Null point is defined as the point between charges where the net force is zero. The electrostatic force at a point can be found by taking the ratio of the product of test charge and given charge at this point to the product of the four-time the value of pi, the permittivity, and the square of distance to this point. Apply the conditions mentioned in the question and substitute the values in it. This will help you in answering this question.

Complete step by step answer:
Let P be the point of zero net force at a distance rr from the charge 11. Let the distance between them be given as,d=2md = 2m. In the region between the 2 charges, the electric field lines will originate at the + charge and terminate at the - charge. Remember that the electric field lines point in the direction of the force on a positive test charge. Therefore the null point of the electric field must lie outside the charges.
q1=+1.10−6{q_1} = + {1.10^{ - 6}}
⇒q2=−4.10−6\Rightarrow {q_2} = - {4.10^{ - 6}}
The electrostatic force at a point can be found by taking the ratio of the product of test charge and given charge at this point to the product of the four-time the value of pi, the permittivity, and the square of distance to this point. Using coulomb's law ,this can be written as,
F=k(q1)(q2)r2F = \dfrac{{k({q_1})({q_2})}}{{{r^2}}}

In this situation, we can write that, as it is already mentioned that the net electrostatic force at the mentioned point is zero. Therefore we assume a test charge qt{q_t} at that point and equate the net force on this test charge to zero. Let the test charge be at the distance rr from the negative charge given. We write electrostatic force due to charge 1 on test charge as,
F−=k(qt)(q2)r2{F_ - } = k\dfrac{{({q_t})({q_2})}}{{{r^2}}}
We write electrostatic force due to charge 2 on test charge as,
F+=k(qt)(q1)(r−2)2{F_ + } = k\dfrac{{({q_t})({q_1})}}{{{{\left( {r - 2} \right)}^2}}}

Now net force will be,
F−+F+=0{F_ - } + {F_ + } = 0
Simplifying this equation can be shown as,
k(qt)(q2)r2+k(qt)(q1)(r−2)2=0k\dfrac{{({q_t})({q_2})}}{{{r^2}}} + k\dfrac{{({q_t})({q_1})}}{{{{\left( {r - 2} \right)}^2}}} = 0
Substituting the values in this equation can be shown as,
(−4×10−6)(r)2+(1×10−6)(r−2)2=0\dfrac{{( - 4 \times {{10}^{ - 6}})}}{{{{\left( r \right)}^2}}} + \dfrac{{(1 \times {{10}^{ - 6}})}}{{{{\left( {r - 2} \right)}^2}}} = 0
⇒1(r−2)2=4(r)2\Rightarrow \dfrac{1}{{{{\left( {r - 2} \right)}^2}}} = \dfrac{4}{{{{\left( r \right)}^2}}}
⇒1(r−2)=2(r)\Rightarrow \dfrac{1}{{\left( {r - 2} \right)}} = \dfrac{2}{{\left( r \right)}}
From this equation, the value of r can be found as,
⇒r=2r−4\Rightarrow r = 2r - 4
∴r=4\therefore r = 4 mm

Therefore the null point is at a distance of 4,m4,m from the negative charge.

Note: Sometimes students might get confused and think that the electric field will be zero at two points. The electric field will be zero at a point nearer to a smaller charge when the system consists of two opposite and unequal charges. The electric field due to a dipole is never zero at any point around the dipole.