# Electric Field Formula

The electric field is formally described as a velocity vector that links the (electromagnetic or Coulomb) pressure of energy imposed over a minuscule single electron at rest at each point in space. Volt/meter (V/m) is the SI unit for electric field strength. The Newtons/coulomb (N/C) unit of electric field strength is also relevant. Electric fields are created by electrostatic force or by the earth’s magnetic field that changes over time.

Rather than thinking of the electrostatic forces as direct contact between different electric currents at a range, one load is thought of as the origin of an electromagnetic current which stretches radially through into the outer atmosphere, and the pressure applied on either a second count in this space is thought of as a directly interacting as between electromagnetic field as well as the second count.

The electromotive force over a solitary positively charged field is directed outward since positive ions resist each other. Electron beams are illustrated as beginning on positive ions and ending on negative charges when portrayed by arcs of power or power traces.

**Formula**

The electric field is denoted by the symbol E. Its dimensional formula is given by the value [M^{1}L^{1}I^{-1}T^{-3}]. It is directly proportional to the force acting on a charge but varies indirectly with the charge value. In other words, its formula equals the ratio of force on a charge to the value of that charge.

E = F/qWhere,

- F is the force acting on the given charge,
- q denotes the charge

**Sample Problems**

**Problem 1: Find the electric field intensity if a force of 10 N acts upon a charge of 8 × 10 ^{-6} C.**

**Solution:**

Given: F = 10 N and q = 8 × 10

^{-6}CSince, E = F/q

= 10/(8 × 10

^{-6})= 1.25 × 10

^{3}N/C

**Problem 2: Find the electric field intensity if a force of 20 N acts upon a charge of 4 × 10 ^{-6} C.**

**Solution:**

Given: F = 20 N and q = 4 × 10

^{-6}CSince, E = F/q

= 20/(4 × 10

^{-6})= 5 × 10

^{6}N/C

**Problem 3: Find the electric field intensity if a force of 20 N acts upon a charge of 7 × 10 ^{-6} C.**

**Solution:**

Given: F = 20 N and q = 4 × 10-6 C

Since, E = F/q

= 20/(4 × 10

^{-6})= 5 × 10

^{6}N/C

**Problem 4: Find the force acting on a charge if the field value is 0.05 N/C and the charge value is 100 C.**

**Solution:**

Given: E = 0.05 N/C and q = 100 C

Since, E = F/q

F = 0.05 (100)

= 5 N

**Problem 5: Find the force acting on a charge if the field value is 1.33 × 10 ^{12} N/C and the charge value is 9 x 10^{-12} C.**

**Solution:**

Given: E = 1.33 × 10

^{-6}N/C and q = 9 × 10^{-6}CSince, E = F/q

F = (1.33 × 10

^{12}) (9 × 10^{-12})= 12 N

**Problem 6: Find the charge if the force acting on it is 20 N and the field value is 4 N/C.**

**Solution:**

Given: F = 20 N and E = 4

Since, E = F/q

q = 20/4

= 5 C

**Problem 7: Find the charge if the force acting on it is 26 N and the field value is 13 N/C.**

**Solution:**

Given: F = 26 N and E = 13

Since, E = F/q

q = 26/13

= 2 C

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