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If a charge of 5 C experiences a force of 25 N in an electric field, the intensity of the field is:
5 N/C
125 N/C
0.2 N/C
20 N/C
5 N/C
The electric field intensity (E) at a point is defined as the force (F) experienced per unit positive charge (q) placed at that point. By dividing the force of 25┬аN by the charge of 5┬аC, we obtain 5┬аN/C.
The electric field intensity (E) at a point is defined as the force (F) experienced per unit positive charge (q) placed at that point. By dividing the force of 25┬аN by the charge of 5┬аC, we obtain 5┬аN/C.
Think of electric field intensity like wind speed: just as wind speed determines the force felt by a sail of a certain size, the electric field intensity determines the force felt by a specific charge.
Remember 'F-E-q' (Force Equals Electric field times charge) or 'E = F/q' (Every Friday, Force over Quality).
E=qFтАЛ тАФ Electric field intensity formula where F is force and q is charge.
The principle is based on the definition of an electric field as a vector field surrounding an electrically charged particle. The magnitude of the field E at any point is determined by the ratio of the electrostatic force exerted on a test charge to the magnitude of that charge, expressed as E=qFтАЛ. This relationship ensures that the field is independent of the test charge magnitude, serving as a measure of the force-exerting capability of the field per unit charge.
The SI unit of electric field intensity is Newtons per Coulomb (N/C) or Volts per meter (V/m).
Electric field intensity is a vector quantity, possessing both magnitude and direction.
The test charge used to measure the field should be small enough not to disturb the existing electric field.
Provides a simple way to calculate force on any charge placed within a known field region.
Does not account for non-uniform fields where the intensity varies with position.
Calculating electron acceleration in cathode ray tubes.
Understanding the behavior of dielectric materials in capacitors.
The electric field direction is defined as the direction in which a positive test charge would move.
Option B (125┬аN/C) results from multiplying instead of dividing (F├Чq).
Option C (0.2┬аN/C) results from an incorrect reciprocal calculation (q/F).
A is correct тАФ The electric field intensity is the ratio of force to charge, calculated as 5┬аC25┬аNтАЛ=5┬аN/C.
Always check units in physics problems; CoulombsNewtonsтАЛ immediately points to the formula E=qFтАЛ.