محاضرات الكهربية الساكنة

(1) What potential difference is needed to stop an electron with an initial speed of 4.2×105m/s? حيث ان الشعل يساوي التغير في طاقة الحركة W = Δk = kf-ki W = 0-ki qV = -1/2 mv2 V = -(1/2 mv2)/q V = -(1/2 x 9.1×10-31x(4.2×105)2)/1.5×10-19 V = -0.57 volt (2) An ion accelerated through a […]

(1) An electric field of intensity 3.5´103N/C is applied the x-axis.  Calculate the electric flux through a rectangular plane 0.35m wide and 0.70m long if (a) the plane is parallel to the yz plane, (b) the plane is parallel to the xy plane, and (c) the plane contains the y axis and its normal makes

(1) The electric force on a point charge of 4.0mC at some point is 6.9´10-4N in the positive x direction.  What is the value of the electric field at that point? من معطيات المسألة (القوة الكهربية الموؤثرة على شحنة في الفراغ) يمكن حساب المجال الكهربي عند تلك النقطة من خلال التعريف الاساسي للمجال وهو حاصل

Example 6.1 An air-filled capacitor consists of two plates, each with an area of 7.6cm2, separated by a distance of 1.8mm.  If a 20V potential difference is applied to these plates, calculate, (a)                the electric field between the plates, (b)               the surface charge density, (c)                the capacitance, and (d)               the charge on each plate. Solution

Example 5.1 What is the potential at the center of the square shown in figure 5.9? Assume that q1= +1 x10-8C,  q2= -2×10-8C,  q3=+3×10-8C, q4=+2×10-8C, and a=1m. Solution   The distance r for each charge from P is 0.71m Example 5.2 Calculate the electric potential due to an electric dipole as shown in figure 5.10. Figure

Example 4.1 If the net flux through a gaussian surface is zero, which of the following statements are true? 1)      There are no charges inside the surface. 2)      The net charge inside the surface is zero. 3)      The electric field is zero everywhere on the surface. The number of electric field lines entering the surface

Example 3.1 Find the electric field at point p in figure 3.4 due to the charges shown. Solution Ex = E1 – E2  =  -36×104N/C Ey = E3 = 28.8×104N/C  Ep = Ö(36×104)2+(28.8×104)2 = 46.1N/C θ = 141o Figure 3.5 Shows the resultant electric field Example 3.2 Find the electric field due to electric dipole

Example 2.2 In figure 2.4, two equal positive charges q=2×10-6C interact with a third charge Q=4×10-6C.  Find the magnitude and direction of the resultant force on Q. Solution لإيجاد محصلة القوى الكهربية المؤثرة على الشحنة Q نطبق قانون كولوم لحساب مقدار القوة التي تؤثر بها كل شحنة على الشحنة Q.  وبما أن الشحنتين q1&q2متساويتان وتبعدان

تعلمنا في الفصول السابقة كيف يمكن التعبير عن القوى الكهربية أو التأثير الكهربي في الفراغ المحيط بشحنة أو أكثر باستخدام مفهوم المجال الكهربي.  وكما نعلم أن المجال الكهربي هو كمية متجهة وقد استخدمنا لحسابه كلا من قانون كولوم وقانون جاوس.  وقد سهل علينا قانون جاوس الكثير من التعقيدات الرياضية التي واجهتنا أثناء إيجاد المجال الكهربي

4.6 Conductors in electrostatic equilibrium A good electrical conductor, such as copper, contains charges (electrons) that are free to move within the material.  When there is no net motion of charges within the conductor, the conductor is in electrostatic equilibrium. Conductor in electrostatic equilibrium has the following properties: Any excess charge on an isolated conductor