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Интензитет на електрическото поле Интензитетът е основна векторна характеристика на електрическото поле; равна на отношението на силата, действаща на точков електрически заряд в дадена точка от полето, към големината на заряда. За да определим електрическото поле в определена точка на пространството, измерваме силата която това поле упражнява върху заряд , когато е поставен в тази точка. се нарича интензитет (напрегнатост) на електрическото поле. В учебниците по английски ще го срещнем само като електрическо поле Е. Да вземем заряд q поставен на разстояние r от тестов заряд . Съгласно закона на Кулон: Съгласно формула (1) Откъдето следва, че: - единичен вектор - определящ посоката на полето. Когато имаме електрическо поле, създадено от множество заряди, интензитета на полето е векторна сума от електрическите полета на всички заряди. -разстояние от заряда qi до точката в която измерваме електрическото поле. Единицата за интензитет е нютон на кулон.
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Obtaining the value of the Electric Field from the Electric Potential The electric field E and the electric potential V are related as shown in Equation 25.3. We now show how to calculate the value of the electric field if the electric potential is known in a certain region. From Equation 25.3 we can express the potential difference dV between two points a distance ds apart as If the electric field has only one component Ex , then E. ds = Ex dx Therefore, Equation 25.15 becomes dV = –Ex dx, or That is, the magnitude of the electric field in the direction of some coordinate is equal to the negative of the derivative of the electric potential with respect to that coordinate. Recall from the discussion following Equation 25.8 that the electric potential does not change for any displacement perpendicular to an electric field. This is consistent with the notion that equipotential surfaces are perpendicular to the field, as shown in Figure 25.12. A small positive charge placed at rest on an electric field line begins to move along the direction of E because that is the direction of the force exerted on the charge by the charge distribution creating the electric field (and hence is the direction of a). Because the charge starts with zero velocity, it moves in the direction of the change in velocity— that is, in the direction of a. In Figures 25.12a and 25.12b, a charge placed at rest in the field will move in a straight line because its acceleration vector is always parallel to its velocity vector. The magnitude of v increases, but its direction does not change. The situation is different in Figure 25.12c. A positive charge placed at some point near the dipole first moves in a direction parallel to E at that point. Because the direction of the electric field is different at different locations, however, the force acting on the charge changes direction, and a is no longer parallel to v. This causes the moving charge to change direction and speed, but it does not necessarily follow the electric field lines. Recall that it is not the velocity vector but rather the acceleration vector that is proportional to force. If the charge distribution creating an electric field has spherical symmetry such that the volume charge density depends only on the radial distance r, then the electric field is radial. In this case, E. ds = Er dr, and thus we can express dV in the form dV = –Er dr, Therefore, For example, the electric potential of a point charge is V = keq/r . Because V is a function of r only, the potential function has spherical symmetry. Applying Equation 25.17, we find that the electric field due to the point charge is Er = keq/r2, a familiar result. Note that the potential changes only in the radial direction, not in any direction perpendicular to r. Thus, V (like Er) is a function only of r. Again, this is consistent with the idea that equipotential surfaces are perpendicular to field lines. In this case the equipotential surfaces are a family of spheres concentric with the spherically symmetric charge distribution (Fig. 25.12b). The equipotential surfaces for an electric dipole are sketched in Figure 25.12c. When a test charge undergoes a displacement ds along an equipotential surface, then dV=0 because the potential is constant along an equipotential surface. From Equation 25.15, then, dV = –E.ds = 0; thus, E must be perpendicular to the displacement along the equipotential surface. This shows that the equipotential surfaces must always be perpendicular to the electric field lines. In general, the electric potential is a function of all three spatial coordinates. If V(r) is given in terms of the cartesian coordinates, the electric field components Ex , Ey , and Ez can readily be found from V(x, y, z) as the partial derivatives For example, if V = 3x2y + y2 + yz, then