Question

What is the strength and direction of the electric field 0.16 m on the left hand side of a 6.7mC positive​

Answer 1

Answer:

you have to specify medium as it is not mentioned here let us take it as vacuum

Explanation:

6.7*10^-6/(.16)^2 *4π€

Answer 2

Explanation:

Given :

Distance of the point on which electric field has to be calculated(r) = 0.16 m

Value of the Charge (q) =

$6.7mC = 6.7 \times {10}^{ - 3} C$

To be Calculated :

Strength and Direction of Electric Filed

Formula to be used :

$E = \frac{kq}{ {r}^{2} } where \: k \: = 9 \times {10}^{9}$

Now,

Calculations :

$E = \frac{kq}{ {r}^{2} } = \frac{9 \times {10}^{9} \times 6.7 \times {10}^{ - 3} }{0.16} = 2.35 \times {10}^{10}$

Hence, the strength of Electric field is 2.35 × 10^10 N/C.

Direction of the Electric Field will be Radially Outwards.

• The electric force per unit charge is referred to as the electric field.
• It is assumed that the field's direction corresponds to the force it would apply to a positive test charge.
• From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.
• Here, since the charge is positive, it's direction will be radially outwards.

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