A charge Q is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is E1 and the intensity at the mid point of the side of square is E2, then the ratio of (E1/E2) will be

Question

A charge Q is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is E1 and the intensity at the mid point of the side of square is E2, then the ratio of (E1/E2) will be (A) (1/2 √2) (B) √2 (C) (1/2) (D) 2

Tardigrade Admin · Accepted Answer

E1=(1/4 π ε0) (Q2/l2) ...(i) E2=(1/4 π ε0) (Q4/l2) E2=2 E1 (E1/E2)=(1/2)

Q. A charge $Q$ is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is $E_{1}$ and the intensity at the mid point of the side of square is $E_{2}$ , then the ratio of $\frac{E _{1}}{E _{2}}$ will be

$\frac{1}{2 2}$

$2$

$\frac{1}{2}$

$2$

$E_{1} = \frac{1}{4 π ε _{0}} \frac{Q ^{2}}{l ^{2}} ... (i)$
$E_{2} = \frac{1}{4 π ε _{0}} \frac{Q ^{4}}{l ^{2}}$
$E_{2} = 2 E_{1}$
$\frac{E _{1}}{E _{2}} = \frac{1}{2}$

Q. A charge Q is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is E1​ and the intensity at the mid point of the side of square is E2​, then the ratio of E2​E1​​ will be

22​1​

2​

21​

2

E1​=4πε0​1​l2Q2​...(i) E2​=4πε0​1​l2Q4​ E2​=2E1​ E2​E1​​=21​

Q. A charge $Q$ is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is $E_{1}$ and the intensity at the mid point of the side of square is $E_{2}$ , then the ratio of $\frac{E _{1}}{E _{2}}$ will be

$\frac{1}{2 2}$

$2$

$\frac{1}{2}$

$2$

$E_{1} = \frac{1}{4 π ε _{0}} \frac{Q ^{2}}{l ^{2}} ... (i)$
$E_{2} = \frac{1}{4 π ε _{0}} \frac{Q ^{4}}{l ^{2}}$
$E_{2} = 2 E_{1}$
$\frac{E _{1}}{E _{2}} = \frac{1}{2}$