Two print charges q-1 - -2-40 n C and q-2 - -6-50 nC are 0-100 m apart- Point A is midway between them - point B is 0-080 m form q-1 and 0-060 m from q-2 as shown in fig- take the electric potential to be zero infinity -The potential point A is

Question

Two print charges  q-1 - -2-40 n C  and  q-2 - -6-50 nC  are 0-100 m apart- Point A is midway between them - point B is 0-080 m form q-1  and 0-060 m from  q-2  as shown in fig- take the electric potential to be zero infinity -The potential point A is

Toppr Admin · Accepted Answer

Correct option is A- -738 VElectric Potential-At - A-160- - V-A - k -160- -dfrac -q-1-r-1- - -dfrac -q-2-r-2- k- -dfrac - 2-40 -times 10-9- C - 0-05 m - - -dfrac - -6-50 -times 10-9- C- -0-05 m-160- - - - -738 V -

Two print charges $$ q_1 = +2.40 n C $$ and $$ q_2 = -6.50 nC $$ are 0.100 m apart. Point A is midway between them ; point B is 0.080 m form $$q_1 $$ and 0.060 m from $$ q_2 $$ as shown in fig. take the electric potential to be zero at infinity .
The potential at point A is

Correct option is A. -738 V
Electric Potential:
At $$ A : V_A = k ( \dfrac {q_1}{r_1} + \dfrac {q_2}{r_2})= k( \dfrac { 2.40 \times 10^{-9} C }{ 0.05 m } + \dfrac { -6.50 \times 10^{-9} C} {0.05 m } ) = -738 V $$

Two print charges $$ q_1 = +2.40 n C $$ and $$ q_2 = -6.50 nC $$ are 0.100 m apart. Point A is midway between them ; point B is 0.080 m form $$q_1 $$ and 0.060 m from $$ q_2 $$ as shown in fig. take the electric potential to be zero at infinity .The potential at point A is

Correct option is A. -738 VElectric Potential:At $$ A : V_A = k ( \dfrac {q_1}{r_1} + \dfrac {q_2}{r_2})= k( \dfrac { 2.40 \times 10^{-9} C }{ 0.05 m } + \dfrac { -6.50 \times 10^{-9} C} {0.05 m } ) = -738 V $$

Two print charges $$ q_1 = +2.40 n C $$ and $$ q_2 = -6.50 nC $$ are 0.100 m apart. Point A is midway between them ; point B is 0.080 m form $$q_1 $$ and 0.060 m from $$ q_2 $$ as shown in fig. take the electric potential to be zero at infinity .
The potential at point A is

Correct option is A. -738 V
Electric Potential:
At $$ A : V_A = k ( \dfrac {q_1}{r_1} + \dfrac {q_2}{r_2})= k( \dfrac { 2.40 \times 10^{-9} C }{ 0.05 m } + \dfrac { -6.50 \times 10^{-9} C} {0.05 m } ) = -738 V $$