What is the strength of the electric field 0.020 m from a 12 µC charge?

Answers

Answer 1
Answer:

The electric field strength at a distance of 0.020 m from a 12 µC point charge is approximately 2.69 x 10⁸ N/C.

What is the electric field strength?

The electric field strength (E) at a point near a point charge is given by the equation:

E = k Q / r²

where:

k is Coulomb's constant, equal to 8.99 x 10⁹ Nm²/C²

Q is the magnitude of the point charge in coulombs

r is the distance from the point charge

Substituting the given values:

Q = 12 x 10⁻⁶ C

r = 0.020 m

E = k Q / r²

E = 8.99 x 10⁹ Nm²/C² x 12 x 10⁻⁶ C / (0.020 m)²

E = 2.69 x 10⁸ N/C

So the electric field strength at a distance of 0.020 m from a 12 µC point charge is approximately 2.69 x 10⁸ N/C.

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Answer 2
Answer: E = ko * C / d^2 = [9.0x10^9 N/m^2C^2 ]* [12x10^-6C]/[(0.020m)^2] = 2.7^10^8N/C

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Chanel has some cotton candy that came in a cloudy shape. She wants to make it more dense. Which describes the candy before and after Chanel manipulated it?

Answers

Answer:

The candy before was fluffy, and the candy after was compacted.

Explanation:

Final answer:

The cloudy cotton candy was light and fluffy due to the air incorporated in it. After Chanel made it dense, the candy became compact and heavier, as she decreased the volume and maintained or increased the amount of sugar.

Explanation:

The cotton candy that Chanel had originally was described as being in a cloudy shape. This suggests that it was light, fluffy, and probably had a lot of air incorporated into it, which is why it was able to maintain such a shape. After she manipulated it to make it more dense, the cotton candy would have lost its cloudy appearance. Instead, it would have become heavier and more compact, because decreasing the amount of air in it increases its density. In other words, the volume of the cotton candy has decreased, but the amount of sugar within that volume has remained the same or increased.

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An acorn with a mass of 0.0300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.00 m/s?

Answers

Kinetic energy = 1/2 (mass) x (speed squared).

KE = 1/2 (0.03) x (5 x 5) = 0.375 kilogram-meter squared / second squared = 0.375 Joule.

The statement "force equals mass times acceleration (F = ma)" is Newton's second law of motion. Why is this a law rather than a theory?Options:
a) It has been extensively tested and confirmed through experimentation.
b) It is a hypothesis that is yet to be proven.
c) It is a general explanation for a wide range of phenomena.
d) It is based on mathematical equations but lacks empirical evidence.

Answers

Answer:

a)  A statement must be tested without contradiction to be a law.

classify the following substances as acidic, basic, or neutral: a. a soapy solution, pH = 9 b. a sour liquid, pH = 5 c. a solution with four times as many hydronium ions as hydroxide ions

Answers

A. Soapy solution (pH of 9) = basic
B. Sour liquid (pH of 5) = acidic
C. Solution with four times as many hydronium (H+) ions as hydroxide ions (OH-) = acidic

Soapy solution with a pH of 9 is basic because it has a pH greater than 7. The same goes for the sour liquid with a pH of 5, which has a pH lower than 7, making it acidic. The word sour also hints at it being acidic because of the definition of Bronsted acids. Finally, higher H+ ion concentrations make a solution more acidic.

A mountain climber ascends a mountain to its peak. The peak is 12,470 ft above sea level. The climber then descends 80 ft to meet a fellow climber. Find the climber's elevation above sea level after meeting the other climber. A. -12,390 ft. B. 12,550 ft. C. 11,670 ft. D. 12,930 ft

Answers

The answer is 12,390 ft.

At first, a climber is at 12,470 ft above sea level. But then, he goes down 80 ft to meet a fellow climber. So, this simply needs to be distracted:
12,470 ft - 80 ft = 12,390 ft
This is the elevation 
above sea level at which he meet the other climber.

with what minimum speed must you toss a 160 g ball straight up to just touch the 13-m-high roof of the gymnasium if you release the ball 1.4 m above the ground? solve this problem using energy.

Answers

energy at top = m*g*height at top from release point
=0.16*9.81*11.6
=18.21J
At release kinetic energy= Gravtiational energy at top
1/2*0.16*v^2=18.21J
v^2=227.625
v=15.10m/s