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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 required to balance?

The intensity of the electric field is given by the ratio of the weight of proton to the charge. So, the magnitude of the electric field is simply defined as **the force per charge on the test charge**.

## 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 and direction of an electric field that will balance?

As the direction of electric force on negatively charged particle is opposite to the direction of electric field, the direction electric filed is **downward**. Hence, the magnitude of electric field is and direction is downwards.

## What is the weight of an electron?

The actual weight of an Electron is **9.05 x 10-28 grams**, that’s right, 9.05 times ten to the minus twenty eighth Grams, an incredibly small mass. Multiply 34.1328 x 1023 times 9.05 x 10-28 and we get 308.9 Grams times ten to minus 5 or in other words, 3.089 Milligrams of electrons.

## What is the electric field inside the capacitor?

Electric field strength

In a simple parallel-plate capacitor, a voltage applied between two conductive plates creates a uniform electric field between those plates. The electric field strength in a capacitor is **directly proportional to the voltage applied and inversely proportional to the distance between the plates**.

## How do you calculate electric field 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=Fq , where E is electric field intensity, F is the force exerted on charge and q is charge.

## 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**.

## What is the magnitude of a point charge which produces an electric field?

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.

## 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.