Electric Field Formula: The electric field (E) at a point in space is defined as the force (F) experienced by a positive test charge (q) placed at that point, divided by the magnitude of the test charge.
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E = -∇V
Where:
- E is the electric field vector. - ∇ (pronounced "nabla") is the del operator, representing the gradient. - V is the electric potential (voltage). In this equation, the negative sign indicates that the electric field points in the direction of steepest decrease in electric potential. In simpler terms, the electric field tells us the direction in which a positive test charge would move if placed in the electric potential field. To put it even more simply, think of electric potential (V) as a hill or slope, where the height of the hill represents the potential energy. The electric field (E) is like the slope of the hill. A positive test charge placed on the hill will roll down the slope in the direction of the steepest descent, which is the direction of the electric field. So, when working with electric potential (V), keep in mind that the electric field (E) is the "slope" or gradient of the potential, and it indicates the direction and magnitude of the force experienced by a charged particle placed in that field. This relationship between electric field and electric potential is a fundamental concept in electromagnetism and is widely used to analyze and understand electric fields in various practical applications, including circuit analysis and the behavior of charged particles in electric fields.Also Check - Soil Erosion Formula
Let's consider a point charge Q = 5 μC placed at a distance of r = 2 meters from a test charge q = 1 μC. Calculate the electric field at this point.
Solution:
Using Coulomb's Law, we can find the force between these charges: F = (k * |Q * q|) / r²Where:
- k is Coulomb's constant (8.99 × 10^9 N m²/C²). F = (8.99 × 10^9 N m²/C² * |5 μC * 1 μC|) / (2 m)² F = (44.95 N m²/C²) / 4 m² F = 11.24 N/C Now, we can find the electric field using the formula E = F / q: E = 11.24 N/C / 1 μC E = 11.24 × 10^6 N/C So, the electric field at this point is 11.24 × 10^6 N/C.Also Check - Efficiency Formula
Example 2: Electric Field Between Two Parallel PlatesConsider two parallel plates with a potential difference of 100 V and a separation of 0.02 meters. Calculate the electric field between the plates.
Solution:
The electric field between parallel plates is given by: E = V / dWhere:
- V is the potential difference. - d is the separation between the plates. E = 100 V / 0.02 m E = 5000 N/C The electric field between the plates is 5000 N/C.