Question

Two like charges of charge +10 nC are placed at the two corners of an equilateral triangle of side 3 cm. What is the magnitude of the total potential at the third corner

Answer 1

Answer:

Using the concept of bond polarity and solubility, explain why water (H2O) and

vinegar are miscible, but vinegar and oil are immiscible.

Answer 2

Answer:

+6μC is the magnitude of the total potential at the third corner.

Explanation:

• In context to the given question, we have to find the magnitude of the total potential at the third corner

⇒ GIVEN:

TWO LIKE CHARGES = +10μC

SIDE OF TRIANGLE = 3cm

⇒SOLUTION:

As we know it the triangle so the distance between charges will be 3cm.

we also know that,

Electric potential (V):- [1/4πε₀]*[q/r]

where; 'q' is charge causing the potential

'r' is distance between the charge and the point where the potential is being measured.

and 'ε₀' is permittivity of free space.

In the given case, the charges are at two vertices of an equilateral triangle and the point where the potential is being measured is the third vertex.

⇒ r =3 cm for both the charges.

The first charge is q₁=+10μC and the second charge is q₂=+10C.

the magnitude of the total potential:

⇒ V=  [1/4πε₀]*[10 x 10⁻⁶/ 3 x 10⁻²] +  [1/4πε₀]*[10 x 10⁻⁶/ 3 x 10⁻²]

⇒ V= 2 { [1/4πε₀]*[10 x 10⁻⁶/ 3 x 10⁻²] }

⇒ V= 2 { [9 x 10⁹] *[10 x 10⁻⁴/ 3] }        [ because [1/4πε₀] = 9 x 10⁹ ]

⇒ V= 2{ [30 x 10⁵] }

⇒ V= 60 x 10⁵ C or +6μC

Therefore, +6μC is the magnitude of the total potential at the third corner.