In the question number 18, the potential at a point 20 cm from the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is

Question

In the question number 18, the potential at a point 20 cm from the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is (A) 1.4 × 105 V (B) 4.2 × 103 V (C) 2.9 × 104 V (D) 3.7 × 105 V

Tardigrade Admin · Accepted Answer

From the figure, potential at P, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8f585df654fdd61f80fa6270086b7d0e-.png" /> V=(q1/4πε0(PA))+(q2/4πε0(PB)) =(1/4πε0(PA))(q1+q2) (∵ PA=PB=√(0.2)2+(0.2)2=0.28 m) =(9×109(1.8+2.8)×10-6/0.28) =1.4×105 V

Q. In the question number $18$ , the potential at a point $20 c m$ from the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is

$1.4 \times 1 0^{5} V$

$4.2 \times 1 0^{3} V$

$2.9 \times 1 0^{4} V$

$3.7 \times 1 0^{5} V$

Q. In the question number 18, the potential at a point 20cm from the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is

1.4×105V

4.2×103V

2.9×104V

3.7×105V

From the figure, potential at P, V=4πε0​(PA)q1​​+4πε0​(PB)q2​​ =4πε0​(PA)1​(q1​+q2​) (∵PA=PB=(0.2)2+(0.2)2​=0.28m) =0.289×109(1.8+2.8)×10−6​ =1.4×105V

Q. In the question number $18$ , the potential at a point $20 c m$ from the mid-point of the line joining the two charges in a plane normal to the line and passing through the mid-point is

$1.4 \times 1 0^{5} V$

$4.2 \times 1 0^{3} V$

$2.9 \times 1 0^{4} V$

$3.7 \times 1 0^{5} V$