A point charge Q = +6.0 \times 10^{-9}\text{ C} is fixed in a vacuum. A small test charge … — JAMB UTME

A point charge $Q = +6.0 \times 10^{-9}\text{ C}$ is fixed in a vacuum. A small test charge $q = +2.0 \times 10^{-9}\text{ C}$ is placed at a distance of $0.30\text{ m}$ from $Q$ . What is the magnitude of the electric field intensity produced by $Q$ at the location of the test charge? [Take $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ ]

A
$1.8 \times 10^2\text{ N C}^{-1}$
$6.0 \times 10^2\text{ N C}^{-1}$ Cevap
C
$8.0 \times 10^2\text{ N C}^{-1}$
D
$1.2 \times 10^{-6}\text{ N C}^{-1}$

Cevap

6.0 \times 10^2\text{ N C}^{-1}

The magnitude of the electric field intensity

E

created by a point charge

Q

at a distance

r

is given by

E = \frac{k Q}{r^2}

. Substituting

k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}

Q = 6.0 \times 10^{-9}\text{ C}

, and

r = 0.30\text{ m}

yields

E = \frac{54}{0.09} = 600\text{ N C}^{-1}

, which equals

6.0 \times 10^2\text{ N C}^{-1}

. The test charge magnitude does not affect the electric field produced by the source charge.

Adım Adım Çözüm

Identify the formula for electric field intensity due to a point charge

E = \frac{k Q}{r^2}

The electric field intensity depends solely on the magnitude of the source charge

Q

and the square of the distance

r

from it.

Substitute the given values into the formula

E = \frac{(9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (6.0 \times 10^{-9}\text{ C})}{(0.30\text{ m})^2}

Insert

k = 9.0 \times 10^9

, source charge

Q = 6.0 \times 10^{-9}\text{ C}

, and distance

r = 0.30\text{ m}

Simplify the arithmetic

E = \frac{54}{0.09} = 600\text{ N C}^{-1} = 6.0 \times 10^2\text{ N C}^{-1}

Squaring

0.30

gives

0.09

, and dividing

54

0.09

yields

600\text{ N C}^{-1}

Anahtar Kavram

Electric Field Intensity of a Point Charge

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