If Q - 2 coloumb and force on it is F - 100 newton- then the value of field intensity will be100 N-C50 N-C200 N-C10 N-C

Question

Toppr Admin · Accepted Answer

The correct option is B 50 N-CAnswer is B-Electric force on a charge q placed in a region o electric field intensity is E and it is given by-xA0-F - qE-In this case- F - 100 N and q - 2 C-So- E-Fq-100N2C-50N-C-Hence-xA0-the value of field intensity will be 50 N-C-

If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be
100 N/C
50 N/C
200 N/C
10 N/C

The correct option is B 50 N/C
Answer is B.

Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, $E = \frac{F}{q} = \frac{100 N}{2 C} = 50 N / C$ .
Hence, the value of field intensity will be 50 N/C.

If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be100 N/C50 N/C200 N/C10 N/C

The correct option is B 50 N/CAnswer is B.Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.In this case, F = 100 N and q = 2 C.So, E=Fq=100N2C=50N/C.Hence, the value of field intensity will be 50 N/C.

If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be
100 N/C
50 N/C
200 N/C
10 N/C

The correct option is B 50 N/C
Answer is B.

Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, $E = \frac{F}{q} = \frac{100 N}{2 C} = 50 N / C$ .
Hence, the value of field intensity will be 50 N/C.