What Will Be The Magnitude Of An Electric Field Intensity That Will Balance The Weight Of An Electron?

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What is the magnitude of electric field intensity that balances the weight of an electron.It is given that e=1. 6×10−19C and me​=9.

What is electric field intensity equal to?

What is Electric Field Intensity? ... The electric field intensity at a point is the force experienced by a unit positive charge placed at that point. Electric Field Intensity is a vector quantity. It is denoted by ‘E’. Formula: Electric Field = F/q.

Which is the magnitude of an electric field intensity that will balance?

What are the magnitude and direction of the electric field that will balance the weight of the following? For an electron/proton we use the equation qE=mg. Therefore the answer for magnitude would just to multiply the mass of an electron/proton by 9.8 .

What is the magnitude of the electric field?

The magnitude of the electric field is simply defined as the force per charge on the test charge . The standard metric units on electric field strength arise from its definition. Since electric field is defined as a force per charge, its units would be force units divided by charge units.

What is the magnitude of the electric field such that an electron placed in the field would experience a force equal to its weight?

Given, The magnitude of electric field intensity E such that an electron placed in it would experience an electrical force equal to its weight is given by ..... therefore electric field intensity E is given by mg/e . option (b) is correct choice.

What is the magnitude of electric field intensity due to a dipole of moment?

In the middle of the dipole, the electric fields contributed by both charges are in the same direction; they are both pointing toward the negative charge. Thus, the magnitude of the electric field is the sum of the contribution of the two charges .

What is the magnitude of a point charge?

The magnitude of the electric field E created by a point charge Q is E=k|Q|r2 E = k | Q | r 2 , where r is the distance from Q. The electric field E is a vector and fields due to multiple charges add like vectors.

What is electric intensity and its unit?

Definition- the electric field intensity is defined as the ratio of the force →F experienced by the charge and the amount of the charge q . ∴→E=→Fq. Force is a vector quantity. The SI unit of the electric field intensity is volt/meter or V/m.

What is the formula of electric intensity?

Hint: The dimensional formula of electric field intensity can be found by using the dimensions of force and charge, as electric field intensity is the force per unit coulomb. Mathematically, $E=dfrac{F}{q}$ , where E is electric field intensity, F is the force exerted on charge and q is charge.

How do you solve electric field?

In vector calculus notation, the electric field is given by the negative of the gradient of the electric potential, E = −grad V . This expression specifies how the electric field is calculated at a given point. Since the field is a vector, it has both a direction and magnitude.

How is electric magnitude calculated?

the magnitude of the electric field (E) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation E = kQ/r 2 , where k is a constant with a value of 8.99 x 10 9 N m 2 /C 2 .

What is K in electric field?

The Coulomb constant, the electric force constant, or the electrostatic constant (denoted k e , k or K) is a proportionality constant in electrostatics equations. In SI units it is equal to 8.9875517923(14)×10 9 kg⋅m 3 ⋅s − 2 ⋅C − 2 .

What is K in Coulomb’s law?

The symbol k is a proportionality constant known as the Coulomb’s law constant. ... Since Coulomb’s law applies to point charges, the distance d in the equation is the distance between the centers of charge for both objects (not the distance between their nearest surfaces).

What is the magnitude of the electric force on an electron in such a field?

Electric Force:

The charge carried by an electron is e=−1.6×10−19 C e = − 1.6 × 10 − 19 C . When a charged particle is placed in an external electric field, it experiences an electric force. the magnitude of this force is equal to the product of charge and electric field .

Which of the following statement is correct the electric field at a point?

Electric field at a point is continuous if there is no charge at that point . And the field is discontinuous if there is charge at that point. So both options (b) and (c) are correct.

Which of the following is deflected by electric field?

X-rays are electromagnetic waves and are not charged. Neutrons are charge-less. Alpha particles are basically helium nuclei and are charged. So they are deflected in the electric field.

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Emily Lee
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