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At a point a short distance from a 4.60x10-6 C charged object, there is an electric field strength of 2.75x105 N/C. What is the distance to the charged object producing this field?

User Turch
by
7.7k points

1 Answer

3 votes

Answer:

0.388

Step-by-step explanation:

To find the distance to the charged object producing the given electric field, we can use Coulomb's law, which relates the electric field strength (E) to the charge (Q) and distance (r) between the charged object and the point where the electric field is measured:

=

2

E=

r

2

k⋅∣Q∣

where:

E is the electric field strength (given as 2.75x10^5 N/C).

k is Coulomb's constant, approximately equal to 8.99x10^9 N m^2/C^2.

∣Q∣ is the magnitude of the charge (given as 4.60x10^-6 C).

r is the distance between the charged object and the point where the electric field is measured (to be found).

Now, let's rearrange the equation to solve for

r:

=

r=

E

k⋅∣Q∣

Substitute the given values:

=

(

8.99

×

1

0

9

N m

2

/

C

2

)

(

4.60

×

1

0

6

C

)

2.75

×

1

0

5

N/C

r=

2.75×10

5

N/C

(8.99×10

9

N m

2

/C

2

)⋅(4.60×10

−6

C)

=

8.99

×

1

0

9

×

4.60

×

1

0

6

2.75

×

1

0

5

r=

2.75×10

5

8.99×10

9

×4.60×10

−6

=

41.434

×

1

0

3

2.75

×

1

0

5

r=

2.75×10

5

41.434×10

3

=

0.15087

r=

0.15087

0.388

meters

r≈0.388meters

The distance to the charged object producing the electric field is approximately 0.388 meters.

User Massy
by
8.5k points