Question

Two positive charges of magnitude 2 and 3 coulombs are placed 10cm apart. The electric potential at a distance of 10cm from the middle point on the right bisector of the line joining the two charges is:
A 5×10^11V
B 4×10^9V
C 4×10^11V
D 5×10^9V
Plz explanation with solution...

Answer 1

net potential at 10 cm distance will be
=potential due to 2coulomb+(potential due to 3coulomb)*2
=9*10power9(2/.15+2*3/.05)
=9*10power9(2/.15 + 6/.05)
=1.2*10power12
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Answer 2

Given: q₁ = 2C

            q₂ = 3C

             distance between the 2 charges = 10cm

             perpendicular distance from middle of the line joining the 2     charges = 10cm

To Find:

Electric potential V = ?

Solution:

Formula used:

• V = kq/r

• H = $\sqrt{Base^{2} + Perpendicular^{2} }$

H or hypotenuse is the distance of charges from the perpendicular point where V is to be found.

base(B) = 5cm

perpendicular(P) = 10cm

r = $\sqrt{B^{2}+ P^{2} }$

r = $\sqrt{2^{2} +3^{2} }$

r = $\sqrt{4 + 9}$

r = $\sqrt{13}$

V = k/r [ q₁ + q₂] (k = 9 × 10⁹)

V = 9 × 10⁹/$\sqrt{13}$[2 + 3]

V = 9 × 10⁹ × 5/$\sqrt{13\\}$

V = 10⁹ × 45/ 3.6

V = 12.5 × 10⁹V