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There are two charges, <br> `q_(1)=5xx10^(-8)C` <br> `q_(2)=-3xx10^(-8)C` <br> Distance between the two charges, d=16 cm =0.16 m <br> consider a point p on the line joining the two charges, as shown in the figure. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_001_S01.png" width="80%"> <br> r=distance of point p from charge `q_(1)` <br> let the electric potential (V) at point p be zero. <br> potential at point p is the sum of potentials caused by charges `q_(1)` and `q_(2)` respectively. <br> `:. V=(q_(1))/(4pi in_(0)r)+(q_(2))/(4pi in_(0)(d-r))......(i)` <br> Where, <br> `in_(0)`=Permittivity of of free space <br> for V=0, equation (i) reduces to <br> `(q_(1))/(4pi in_(0)r)=-(q_(2))/(4pi in_(0)(d-r))` <br> `(q_(1))/r=(-q_(2))/(d-r)` <br> `0.16/4 -1=3/5` <br> `0.16/r =8/5` <br> `:. r=0.1 m =10 cm ` <br> Therefore, the potential is zero at distance of 10 cm from the positive charge between <br> the charges. <br> Supposite point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in figure in the following figure. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_001_S02.png" width="80%"> <br> for this arragements , potential is given by <br> `V=(q_(1))/(4pi in_(0)s)+(q_(2))/(4pi in_(0)(s-d)) ......(ii)` <br> For v=0 , equation (ii) reduces to <br> `(q_(1))/(4piin_(0)s) =-(q_(2))/(4piin_(0)(s-d))` <br> `(q_(1))/s=(-q_(2))/(s-d)` <br> `(5xx10^(-8))/s=((-3xx10^(-8)))/((s-0.16))` <br> `1-0.16/s=3/5` <br> `0.16/s=2/5` <br> Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.Related Video

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