A force of 2-25N acts on a charge of 15times -10-4-C- The intensity of electric field that point is then150Nquad - C - -1 -15Nquad - C - -1 -quad quad 1500Nquad - C - -1 -1-5Nquad - C - -1 -

Question

Toppr Admin · Accepted Answer

Electric field- E-Fq-2-25N15-xD7-10-x2212-4C-1500NC-x2212-1

A force of $2.25 N$ acts on a charge of $15 \times 10^{- 4} C$ . The intensity of electric field at that point is then
$150 N C^{- 1}$
$15 N C^{- 1}$
$1500 N C^{- 1}$
$1.5 N C^{- 1}$

Electric field, $E = \frac{F}{q} = \frac{2.25 N}{15 \times 10^{- 4} C} = 1500 N C^{- 1}$

A force of 2.25N acts on a charge of 15×10−4C. The intensity of electric field at that point is then150NC−115NC−11500NC−11.5NC−1

Electric field, E=Fq=2.25N15×10−4C=1500NC−1

A force of $2.25 N$ acts on a charge of $15 \times 10^{- 4} C$ . The intensity of electric field at that point is then
$150 N C^{- 1}$
$15 N C^{- 1}$
$1500 N C^{- 1}$
$1.5 N C^{- 1}$

Electric field, $E = \frac{F}{q} = \frac{2.25 N}{15 \times 10^{- 4} C} = 1500 N C^{- 1}$