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 B 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 B is

Toppr Admin · Accepted Answer

Correct option is C- -705 VElectric Potential-At - B - V-B - k - -dfrac -q-1-r-1- -dfrac -q-2-r-2- - k-dfrac -2-40 -times 10-9- C-0-08m- - -dfrac - -6-50 -times 10-9- C-0-06 m- - - -705 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 B is

Correct option is C. -705 V
Electric Potential:
At $$ B : V_B = k ( \dfrac {q_1}{r_1} +\dfrac {q_2}{r_2})$$
$$ = k(\dfrac {2.40 \times 10^{-9} C}{0.08m} + \dfrac { -6.50 \times 10^{-9} C}{0.06 m} ) = -705 V $$

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 B is

Correct option is C. -705 VElectric Potential:At $$ B : V_B = k ( \dfrac {q_1}{r_1} +\dfrac {q_2}{r_2})$$$$ = k(\dfrac {2.40 \times 10^{-9} C}{0.08m} + \dfrac { -6.50 \times 10^{-9} C}{0.06 m} ) = -705 V $$

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 B is

Correct option is C. -705 V
Electric Potential:
At $$ B : V_B = k ( \dfrac {q_1}{r_1} +\dfrac {q_2}{r_2})$$
$$ = k(\dfrac {2.40 \times 10^{-9} C}{0.08m} + \dfrac { -6.50 \times 10^{-9} C}{0.06 m} ) = -705 V $$