Question

Charges of 6,12 & 24 micro microcoulomb of +ve electricity are placed at the corners of a square. Find what charge must be placed at the fourth corner, so that the potential at the centre of the square may be zero.​

Answer 1

Question :-

Positive charges of 6,12 and 24 μC of positive charge are placed at the three vertices of a square. What charge must be placed at fourth vertex so that total potential at the center of the square is zero?

Solution :-

• Charge at corner A = 6 μC

• Charge at corner B = 12 μC

• Charge at corner C = 24 μC

• Charge at corner D = x μC                  ...(To be found)

The distance from the center of the square to all its vertices are equal.

Let the square be ABCD.

⇒ Let the center be the point where its diagonals intersect.

⇒ The distance from that point to all the vertices is equal.

$\sf{V_C_e_n_t_e_r\ by\ A=\frac{k(6)}{r} }$

$\sf{V_C_e_n_t_e_r\ by\ B=\frac{k(12)}{r} }$

$\sf{V_C_e_n_t_e_r\ by\ C=\frac{k(24)}{r} }$

Let the charge of D be x.

Then,

$\sf{V_C_e_n_t_e_r\ by\ D=\frac{k(x)}{r} }$

When we add all these, the potential becomes 0.      (Given)

$\sf{V_t_o_t_a_l=\frac{k}{r}(6+12+24+x)=0 }\\\\\sf{\implies 6+12+24+x=0}\\\\\sf{\implies 42+x=0}\\\\\boxed{\sf{\implies x=-42}}$

∴So, the charge at the fourth corner D is -42 μC for which the potential at the center of the square will be zero.

Answer 2

Explanation:

hope it was helpful for you to solve problem